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RESEARCHES 



IN 



GRAPHICAL STATICS 



RESEARCHES 



IN 



GRAPHICAL STATICS. 



BY 



HENRY T. EDDY, C. E., Ph. D., 

Professor of Mathematics and Civil Engineering 
IN the University of Cincinnati. 



Illustrated by Forty-one Engravings in Text and Nine Folding Plates. 



REPRINTED FROM VAN NOSTRAND' S ENGINEERING MAGAZINE. 



Vr _ 



NEW YORK: 

U). VAN NOSTHAND, f»u.blisher, 

23 MuKiiAY Street and 27 Wakuen Street. , 

1 S78. 






Copyright ; 

1878, 

By D. Vah Nostkand. 









(^ 





\> 



PREFACE. 



At a meeting of the American Association for the Advancement of Science, 
held in August, 18*76, at Buffalo, the writer read two papers, entitled respect- 
ively, "Certain New Constructions in Graphical Statics," and "A New Funda- 
mental Method in Graphical Statics." These papers, with considerable addi- 
tions and amplifications, are presented on the following pages ; and to them 
is added a third on The Theory of Internal Stress. 

The paper, entitled JVew Constructions in Graphical /Statics, is largely- 
occupied with the various forms of the elastic arch. The possibility of obtain- 
ing a complete graphical solution of the elastic arch in all cases depends upon 
a theorem not hitherto recognized as to the relative position of the equilibrium 
curve due to the loading and the curve of the arch itself. The demonstration 
of this theorem, which may be properly named the Theorem Respecting the 
Coincidence of Closing Lines, as given on page 12, is somewhat obscure. How- 
ever, a second demonstration is given on page 98, and this latter, stated at 
somewhat greater length, may also be found in the America?! Journal of Pure 
and Applied Mathematics, Vol. I, No. 3. Prof. Wm. Cain, A.M., C.E., has 
also published a third demonsiration in Van Nostrand''s Magazine, Vol. XVIII. 
The solution of the elastic arch is further simplified so that it depends upon that 
of the straight girder of the same cross section. Moreover, it is shown that 
the processes employed not only serve to obtain the moment, thrust and shear 
due the loading, but also to obtain those due to changes of temperature, or to 
any cause which alters the span of the arch. It is not known that a graphical 
solution of temperature stresses has been heretofore attempted. 

A new general theorem is also enunciated which affords the basis for a 
direct solution of the flexible arch rib, or suspension cable, and its stiffening 
truss. 

These discussions have led Jo a new graphical solution of the continuous 
girder in the most general case of variable moment of inertia. This is accom- 
panied by an analytic investigation of the Theorem of Three Moments, in 
which the general equation of three moments appears for the first time in 
simple form. This investigation, slightly extended and amplified, may be also 
found in the American Journal of Pure and Applied Mathematics, Yol. I, No. 1. 

Intermediate between the elastic and flexible arch is the arch with block- 
work joints, such as are found in stone or brick arches. A graphical solution 
of this problem was given by Poncelet, which may be found in Woodbury's 
treatise on the Stability of the Arch, page 404. Woodbury states that this 
solution is correct in case of an unsymmetrical arch, but in this he is mis- 
*taken. The solution proposed in the following pages is simpler, susceptible 



VI PREFACE. 



of greater accuracy, and is not restricted to the case when either the arch or 
loading is symmetrical about the crown. 

The graphical construction for determining the stability of retaining walls 
is the first one proposed, so far as known, which employs the true thrust in 
its real direction, as shown by Rankine in his investigation of the stress of 
homogeneous solids. It is in fact an adaptation of that most useful conception, 
Coulomb's Wedge of Maximum Thrust, to Rankine's investigation. 

It has also been found possible to obtain a complete solution of the dome 
of metal and of masonry by employing constructions analogous to those em- 
ployed for the arch ; and in particular, it is believed that the dome of masonry 
is here investigated correctly for the first time, and the proper distinctions 
pointed out between it and the dome of metal. 

In the paper entitled, A New General Method in Graphical Statics, a 
fundamental process or method is established of the same generality as the 
well-known method of the Equilibrium Polygon. The new method is designated 
as that of the Frame Pencil, and both the methods are discussed side by side 
in order that their reciprocal relationship may be made the more apparent. The 
reader who is not familiar with the properties of the equilibrium polygon will 
find it advantageous to first read this paper, or, at least, defer the others until 
he has read it as far as page 83. 

As an example affording a comparison of the two methods, the moments of 
inertia and resistance have been discussed in a novel manner, and this is ac- 
companied by a new graphical discussion of the distribution of shearing stress. 

In the paper entitled. The Theory of Internal Stress in Graphical Statics, 
there is considerable new matter, especially in those problems which relate to 
the combination of states of stress, a subject which has not been, heretofore, 
sufficiently treated. 

It is hoped that these graphical investigations which afford a pictorial repre- 
sentation, so to speak, of the quantities involved and their relations may not 
present the same difficulties to the reader as do the intricate formulae arising 
from the analytic solutions of the same problems. Indeed, analysis almost 
always requires some kind of uniformity in the loading and in the structure 
sustaining the load, while a graphical construction treats all cases with the 
same ease ; and especially are cases of discontinuity, either in the load or 
structure, difficult by analysis but easy by graphics. 

H. T. E. 



ooiSTTE r<nrs. 



NEW CONSTRUCTIONS IN aRAPHICAL STATICS. 

PAGE. 

The Fundamental Propositions and Equations of the Elastic Arch 9 

Arch Rib with Fixed Ends 14 

Arch Rib with Fixed Ends and Hinge Joint at the Crown 20 

Temperature Strains 23 

Unsymmetrical Arches 25 

Arch Rib with End Joints 25 

Arch Rib with Three Joints , 27 

Arch Rib with One End Joint 28 

Arch Rib with Two Joints 29 

Suspension Bridge with Stiffening Truss 30 

Continuous Girder with Variable Cross Section 36 

Theorem of Three Moments 41 

Flexible Arch Rib and Stiffening Truss 43 

Arch of Masonry 45 

Retaining Walls and Abutments 49 

Foundations in Earth = 53 

Spherical Dome of Metal 53 

Spherical Dome of Masonry 56 

Conical Dome of Metal 59 

Conical Dome of Masonry 60 

Other Vaulted Structures 61 

A NEWJGENERAL METHOD IN GRAPHICAL STATICS. 

Reciprocal Figures 65 

Roof Truss, Wind and Weight Stresses 67 

Bridge Truss, Maximum Stresses 68 

Wagon Wheel with Tension-Rod Spokes 72 



Vm CONTENTS. 



PAGE. 

Water Wheel with Tension-Rod Spokes 73 

Equilibrium Polygan for Forces in One Plane 74 

Frame Pencil for Forces in One Plane 76 

Equilibriilm Forces for Parallel Forces 78 

Frame Pencil for Parallel Forces 80 

Summation Polygon 82 

Girder with Fixed Ends, by both Methods 83 

Any Forces in One Plane applied at Given Points 86 

Kernel, Moments of Resistance and Inertia, treated by the Equilibrium 

Polygon 86 

Kernel, Moments of Resistance and Inertia, treated by the Frame Pencil 90 

Uniformly Varying Stress in General 91 

Distribution of Shearing Stress 92 

Relative Stresses 94 

Stresses in a Horizontal Chord 97 

Addendum to Page 12 98 

Addendum to Page 10 99 

t 

THE THEOEY OF INTEENAL STEESS IN GEAPHIOAL STATICS. 

Introductory Definitions, etc 103 

General Properties of Plane Stress 104 

Problems in Plane Stress 110 

Combination and Separation of States of Stress 116 

Properties of Solid Stress 119 

Problems in Solid Stress 120 



^ I O » ♦ I ^ 



ERRATA. 

Page 12, line 12, first column, for " these " put " their." 

" 42, " 16, " " " {Mi) " (M^i). 



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55, 


" 26, 


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a 


ii 


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" 11, 


second 


ii 


a 


B. C] 


L'emona 


put L. Cremona. 



NEW CONSTRUCTIONS 



IN 



GRAPHICAL STATICS 



NEW CONSTRUCTIONS 



IN 



,GPvAPHICAL STATICS 



CHAPTER I. 

It is the object of this work to fully dis- 
cuss the stability of all forms of the arch, 
flexible or rigid, by means of the equili- 
brium polygon — the now well recognized 
instrument for graphical investigation. 
One or two other constructions of inter- 
est may also be added in the sequel. 
The discussion will presuppose an ele- 
mentary knowledge of the properties of 
the equilibrium polygon, and its accom- 
panying force polygon, for parallel 
forces. 

As ordinarily used in tlie discussion of 
the simple or continuous girder, the 
equilibrium polygon has an entirely arti- 
ficial relation to the problem in hand, 
and the particular horizontal stress as- 
sumed is a matter of no consequence ; 
but not so with respect to the arch. As 
will be seen, there is a special equili- 
brium polygon appertaining to a given 
arch and load, and in this particular 
polygon the horizontal stress is the ac- 
tual horizontal thrust of the arch. W?ien 
this thrust has been found in any given 
case, it permits an immediate determ- 
ination of all other questions respecting 
the stresses. This thrust has to be de- 
termined differently in arches of differ- 
ent kinds, the method being dependent 
upon the number, kind, and position of 
the joints in the arch. 

The methods we shall use depend upon 
our ability to separate the stresses in- 
duced by the loading into two parts; one 



part being sustained in virtue of the re- 
action of the arch in the same manner as 
an inverted suspension cable (^.e., as an 
equilibrated linear archj, and the remain- 
der in virtue of its reaction as a girder. 
These two ways in which the loading is 
sustained are to be considered somewhat 
apart from each other. To this end it 
appears necessary to restate and discuss, 
in certain aspects, the well-known equa- 
tions applicable to elastic girders acted 
on by vertical pressures due to the load 
and the resistances of the supports. 

Let jP represent any one of the various 
pressures, Pj, P^j -^n, applied to the 
girder. 

Consider an ideal cross section of the 
girder at any point 0. 

Let i?=the horizontal distance from O 
to the force P. 

'Let ^=the radius of curvature of the 
girder at 0. 

At the cross section O, the equations 
just mentioned become : — 

Shearing stress, S=^ (P) 

Moment of flexure, J/=^ (Px) 



Curvature, 



~B EI 



Total bending, B=:E{P) = ^ (-^^) 
Deflection, D = :S (Px) = 2 (^^) 



'KhAjV^ 



10 



NEW CONSTRUCTIOlSrS 



in which E is the modulus of elasticity 
of the material, and I is the moment of 
inertia of the girder; and as is well 
known, the summation is to be extended 
from the point to a free end of the 
girder, or, if not to a free end, the sum- 
mation expresses the effect only of the 
quantities included in the summation. 

Let a number of points be taken at 
equal distances along the girder, and let 
the values of P^ S, 3f, J5, D be com- 
puted for these points by taking at 
these points successively, and also erect 
ordinates at these points whose lengths 
are proportional to the quantities com- 
puted. First, suppose T is the same at 
each of the points chosen, then the 
values of these ordinates may be ex- 
pressed as follows, if a, b, c, etc., are any 
real constants whatever : 



yp = a . P . 


(1) 


ys^b.2{P) . . 


(2) 


ym= o.2(Px)=c.M.. 


(3) 


yb = cl.2{M) . 


(4) 


yd = e . :S{Mx) . 


(5) 



If I is not the same at the different 
cross sections, let P=^3f-^I • then the 
last three equations must be replaced 
by the following: 



ym'=.f.P' 



(3') 
(4') 



The ordinates ym and ym' are not 
equal, but can be obtained one from the 
other when we know the ratio of the 
moments of inertia at the different cross 
sections. 

Equation (1) expresses the loading, 
and yp may be considered to be the 
depth of some uniform material as 
earth, shot or masonry constituting the 
load. Lines joining the extremities of 
these ordinates will form a polygon, or 
approximately a curve which is the up- 
per surface of such a load. When the 
load is uniform the surface is a hori- 
zontal line. 

For the purposes of our investiga- 
tion, a distributed load whose upper 



surface is the polygon or curve, above 
described, is considered to have the 
same effect as a series of concentrated 
loads proportional to the ordinates 
yp acting at the assumed points of 
division. If the points of division be 
assumed sufficiently near to each other, 
the assumption is sufficiently accurate. 

If a polygon be drawn in a similar 
manner by joining the extremities of the 
ordinates ym computed from equation 
(3), it is known that this polygon is an 
equilibrium polygon for the applied 
weights P, and it can also be construct- 
ed directly without computation by the 
help of a force polygon having some as- 
sumed horizontal stress. 

Now, it is seen by inspection that 
equations (3) and (5), or (3') and (5'), 
have the same relationship to each other 
that equations (1) and (3) have. The re- 
lationship may be stated thus : — If the 
ordinates ym (or ym') be regarded as 
the depth of some species of loading, so 
that the polygonal part of the equili- 
brium polygon is the surface of such 
load, then a second equilibrium polygon 
constructed for this loading will have for 
its ordinates proportional to yd. But 
these last are proportional to the actual 
deflections of the girder. 

Hence a second equilibrium polygon, 
so constructed, might be called the de- 
flection polygon, as it shows on an ex- 
aggerated scale the shape of the neutral 
axis of the deflected girder. 

The first equilibrium polygon having 
the ordinates ym may be called the mo- 
ment polygon. 

It may be useful to consider the physi- 
cal significance of equations (3), (4), (5), 

or (3'), (4'), (5'). 

According to the accepted theory of 
perfectly elastic material, the sharpness 
of the curvature of a uniform girder is 
directly proportional to the moment of 
the applied forces, and for different 
girders or different portions of the same 
girder, it is inversely proportional to the 
resistance which the girder can afford. 
Now this resistance varies directly as I 
varies, hence curvature varies as M-^I^ 
which is equation (3) or (3'). 

Now curvature, or bending at a point, 
is expressed by the acute angle between 
two tangents to the curve at the distance 
of a unit from each other: and the total 



IN GRAPHICAL STATICS. 



11 



bending, i.e. the angle between the tan- 
gent at O, and that at some distant point 
A is the sum of all such angles between 
O and the point A. Hence the total 
bending is proportional to ^(Jf-f-Z), 
the summation being extended from 
to the ]3oint A, which is equation (4) or 

Again, if bending occurs at a point 
distant from 0, as A, and the tangent at 
A be considered as fixed, then is de- 
flected from this tangent, and the 
amount of such deflection depends both 
upon the amount of the bending at A, 
and upon its distance from 0. Hence 
the deflection from the tangent at A is 
proportional to ^ (Mx~J) which is 
equation (5) or (5').. 

It will be useful to state explicitly 
several propositions, some of which are 
implied in the foregoing equations. The 
importance and applicability of some of 
them has not, perhaps, been sufliciently 
recognized in this connection. 

Prop. I. Any girder (straight or other- 
wise) to which vertical forces alone are 
applied (^. e., there is no horizontal 
thrust) sustains at any cross-section the 
stress due to the load, solely by develop- 
ing one internal resistance equal and op- 
posed to the shearing, and another equal 
and opposed to the moment of the applied 
forces. 

Prop. II. But any flexible cable or 
arch with hinge joints can offer no re- 
sistance at these joints to the moment 
of the applied forces, and their moment 
is sustained by the horizontal thrust de- 
veloped at the supports and by the ten- 
sion or compression directly along the 
cable or arch. 

It is well known that the equilibrium 
polygon receives its name from its being 
the shape which such a flexible cable, or 
equilibrated arch, assumes under the 
action of the forces. In this case we 
may say for brevity, that the forces are 
sustained by the cable or arch in virtue 
of its being an equilibrium polygon. 

Prop. III. If an arch not entirely flexi- 
ble is supported by abutments against 
which it can exert a thrust having a 
horizontal component, then the moment 



due to the forces applied to the arch will 
be sustained at those points which are 
not flexible, partly in virtue of its being 
approximately an equilibrium polygon, 
and partly in virtue of its resistance as a 
girder. 

It is evident from the nature of the 
equilibrium polygon that it is possible 
with any given system of loading to make 
an arch of such form (viz., that of an equi- 
librium polygon) as to require no bracing 
whatever, since in that case there will 
be no tendency to bend at any point. 
Also it is evident that any deviation of 
part of the arch from this equilibrium 
polygon would need to be braced. As, 
for example, in case two distant points 
be joined by a straight girder, it must 
be braced to take the place of part of 
the arch. Furthermore, the greater the 
deviation the greater the bending mo- 
ment to be sustained in this manner. 
Hence appears the general truth stated 
in the proposition. 

It will be noticed that the moment 
called into action, at any point of a straight 
girder, depends not only on the applied 
forces which furnish the polygonal part 
of the equilibrium polygon, but also on 
the resistance which the girder is capa- 
ble of sustaining at joints or supports, or 
the like. For example, if the girder 
rests freely on its end-supports, the mo- 
ment of resistance vanishes at the ends, 
and the "closing line" of the polygon 
joins the extremities of the polygonal 
part. If however the ends are fixed 
horizontally and there are two free 
(hinge) joints at other points of the gir- 
der, the polygonal part will be as before, 
but the closing line would be drawn so 
that the moments at those two points 
vanish. Similarly in every case (though 
the conditions may be more complicated 
than in the examples used for illustration) 
the position of the closing line is fixed 
by the joints or manner of support of 
the girders, for these furnish the condi- 
tions which the moments {i. e., the ordi- 
nates of the equilibrium polygon) must 
fulfill. For example, in a straight uni- 
form girder without joints and fixed 
horizontally at the ends, tlie conditions 
are evidently these; the total bendino- 
vanishes when taken from end to end 
and the deflection of one end below the 
tangent at the other end also vanishes. 



a>^^ 



12 



NEW CONSTEUCTIONS 



Prop. TV. If in any arch that equilibrium 
polygon (due to the weights) be construct- 
ed which has the same horizontal thrust 
as the arch actually exerts; and if its 
closing line be drawn from consideration 
of the conditions imposed by the supports, 
etc. ; and if furthermore the curve of the 
arch itself be regarded as another equilib- 
rium polygon due to some system of load- 
ing not given, and its closing line be also 
found from the same considerations re- 
specting supports, etc., then, when these 
two polygons are placed so that these 
closing lines coincide and their areas 
partially cover each other, the ordinates 
intercepted between these two polygons 
are proportional to the real bending mo- 
ments acting in the arch. 

Suppose that an equilibrium polygon 
due to the weights be drawn having the 
same horizontal thrust as the arch. We 
are in fact unable to do this at the out- 
set as the horizontal thrust is unknown. 
We only suppose it drawn for the pur- 
pose of discussing its properties. Let 
also the closing line be drawn, which 
may be done, as will be seen hereafter. 
Call the area between the closing line 
and the polygon, A. Draw the closing- 
line of the curve of the arch itself (re- 
garded as an equilibrium polygon) ac- 
cording to the same law, and call the 
area between this closing line and its 
curve A". Further let A' be the area of 
a polygon whose ordinates represent the 
actual moments bending the arch, and 
drawn on the same scale as A and A". 
Since the supports etc., must influence 
the position of the closing line of this 
polygon in the same manner as that of 
A, we have by Prop. Ill not only 

A=A'-hA" 

which applies to the entire areas, but 
also 

y=y'-\-y" 

as the relation between the ordinates of 
these polygons at any of the points of 
division before mentioned, from which 
the truth of the proposition api^ears. 

This demonstration in its general form 
may seem obscure since the conditions 
imposed by the supports, etc., are quite 



various, and so cannot be considered in 
a general demonstration. The obscurity, 
however, will disappear after the treat- 
ment of some particular cases, where we 
shall take pains to render the truth of 
the proposition evident. We may, how- 
ever, make a statement which will pos- 
sibly put the matter in a clearer light by 
saying that A" is a figure easily found, 
and we, therefore, employ it to assist in 
the determination of A' which is un- 
known, and of A which is partially un- 
known. And we arrive at the peculiar 
property oiA\ that its closing line is found 
in the same manner as that of ^, by no- 
ticing that the positions of the closing 
lines of A and A' are both determined 
in the same manner by the supports, etc. ; 
for the same law would hold when the 
rise of the arch is nothing as when it 
has any other value. But A" is the dif- 
ference of A and A'. Hence what is 
true of A and A' separately is true of 
their difference A\ the law spoken of 
being a mere matter of summation. 

From this proposition it is also seen that 
the curve of the arch itself may be re- 
garded as the curved closing line of the 
polygon whose ordinates are the actual 
bending moments, and the polygon it- 
self is the polygonal part of the equili- 
brium polygon due to the weights. 

It is believed that Prop. IV contains 
an important addition to our previous 
knowledge as to the bending moments in 
' an arch, and that it supplies the basis 
j for the heretofore missing method of 
I obtaining graphically the true equili- 
j brium polygon for the various kinds of 
! arches. 

Prop. y. If bending moments M act 

on a uniform inclined girder at horizon- 
tal distances x from 0, the amount of the 
vertical deflection yd will be the same 
as that of a horizontal girder of the 
same cross section-, and having the same 
horizontal span, upon which the same 
moments M act at the same horizontal 
distances x from 0. Also, if bending 
moments JIfact as before, the amount of 
the horizontal deflection, say xa, will be 
the same as that of a vertical girder of 
the same cross section, and having the 
same height, upon which the same mo- 
ments M act at the same heights. 



IN GRAPHICAL STATICS. 



18 




cy 



Let the moment M act at A, produc- 
ing according to equation (5) the deflec- 
tion 

0C=e.3f.A0 

whose vertical and horizontal compo- 
nents are 

y^ — CE and xa — OE 

For the small deflections occurring in a 
girder or arch, ^OC=90° 

,\ A0\ 0F\\ OC: CE 

.-. GE^^.OF^e.M.OF 
AO 

,'. q/d = e . Mx 
Also, AO : AF\\ 00: OE 

.'. OE=^,AF=e.M.AF 
AO 

.-. Xd = e.My 

The same may be proved of any other 
moments at other points; hence a simi- 
lar result is true of their sum; which 
proves the proposition. 

It may be thought that the demonstra- 
tion is deficient in rigor by reason of the 
assumption that ^4 0(7=90°. 

Such, however, is not the fact as ap- 
pears from the analytic investigation of 
this question by Wm. Bell in his at- 
tempted graphical discussion of the arch 
in Vol, VIII of this Magazine, in which 
the only approximation employed is that 
admitted by all authors in assuming that 
the curvature is exactly proportional to 
the bending moment. 

We might in this proposition substi- 
tute f. M-^I for e . M, and prove a 
similar but more general proposition re- 



specting deflections, which the reader 
can easily enunciate for himself. 

Before entering upon the particular 
discussions and constructions we have in 
view, a word or two on the general 
question as to the manner in which the 
problem of the arch presents itself, will 
perhaps render apparent the relations 
between this and certain previous inves- 
tigations. The problem proposed by 
Rankine, Yvon-Villarceaux, and other 
analytic investigators of the arch, has 
been this: — Given the vertical loading, 
what must be the form of an arch, and 
what must be the resistances of the 
spandrils and abutments, when the 
weights produce no bending moments 
whatever? By the solution of this ques- 
tion they obtain the equation and prop- 
erties of the particular equilibrium poly- 
gon which would sustain the given 
weights. Our graphical process com- 
pletely solves this question by at once 
constructing this equilibrium polygon. 
It may be remarked in this connection, 
that the analytic process is of too com- 
plicated a nature to be effected in any, 
except a few, of the more simple cases, 
while the graphical process treats all 
cases with equal ease. 

But the kind of solution just noticed, 
is a very incomplete solution of the 
problem presented in actual practice ; 
for, any moving load disturbs the dis- 
tribution of load for which the arch is 
the equilibrium polygon, and introduces 
bending moments. For similar reasons 
it is necessary to stiffen a suspension 
bridge. The arch must then be propor 
tioned to resist these moments. Since 
this is the case, it is of no particular 
consequence that the form adopted for 
the arch in any given case, should be 
such as to entirely avoid bending mo- 
ments when not under the action of the 
moving load. 

So far as is known to us, it is the 
universal practice of engineers to as- 
sume the form and dimensions, as 
well as the loading of any arch pro- 
jected, and next to determine whether 
the assumed dimensions are consistent 
with the needful strength and stability. 
If the assumption is unsuited to the 
case in hand, the fact will appear by the 
introduction of excessive bending mo- 
ments at certain points. The considera- 
tions set forth furnish a guide to a new 



14 



TsTEW COJ^STEUCTIOlSrS 



assuraption which shall be more suitable, 
it being necessary to make the form of 
the arch conform more closely to that of 
the equilibrium polygon for the given 
loading. 

The question may be regarded as one 
of economy of material, and ease of 
construction, analogous to that of the 
truss bridge. In this latter case, con- 
structors have long since abandoned any 
idea of making bridges in which the 
inclination of the ties and posts should 
be such as to require theoretically the 
minimum amount of material. Indeed, 
the amount of material in the case of a 
theoretic minimum, differs by such an 
inconsiderable quantity from that in 
cases in which the ties and posts have a 
very different inclination, that the attain- 
ment of the minimum is of no practical 
consequence. 

Similar considerations applied to the 
arch, lead us to the conclusion that the 
form adopted can in every case be 
composed of segments of one or more 
circles, and that for the purpose of con- 
struction every requirement will then be 
met as fully as by the more complicated 
transcendental curA^es found by the 
writers previously mentioned. If con- 
siderations of an artistic nature render 
it desirable to adopt segments of para- 
bolas, ellipses or other ovals, it will be a 
matter of no more consequence than is 
the particular style of truss adopted by 
rival bridge builders. 

We can also readily treat the problem 
in an inverse manner, viz : — find the 
system of loading, of which the assumed 
curve of the arch is the equilibrium 
polygon. From this it will be known 
how to load a given arch so that there 
shall be no bending moments in it. 
This, as may be seen, is often a very 
useful item of information ; for, by leav- 
ing open spaces in the masonry of the 
spandrils, or by properly loading the 
crown to a small extent, we may fre- 
quently render a desirable form entirely 
stable and practicable. 

CHAPTER II. 

THE ARCH ElB WITH FIXED ENDS. 

Let us take, as the particular case to 
be treated, that of the St. Louis Bridge, 
which is a steel arch in the form of the 



arc of a circle ; having a chord or span 
of 518 feet and a versed sine or rise of 
one-tenth the span, i. e. 51.8 feet. The 
arch rib is firmly inserted in the im- 
mense skew-backs which form part of 
the upper portion of the abutments. It 
will be assumed that the abutments do 
not yield to either the thrust or weight 
of the arch and its load, which was also 
assumed in the published computations 
upon which the arch was actually con- 
structed. Further, we shall for the 
present assume the cross section of the 
rib to have the same moment of iner- 
tia, T, at all points, and shall here only 
consider the stresses induced by an 
assumed load. The stresses due to 
changes in the length of the arch itself, 
due to its being shortened by the load- 
ing, and to the variations of temperature, 
are readily treated by a method similar 
to the one which will be used in this 
article, and will be treated in a subse- 
quent chapter. 

Let ^g a h\' in Fig. 2, be the neutral 
axis of the arch of which the rise is one- 
tenth the span. Let a x y z be the area 
representing the load on the left half of 
the arch, and a x' y' z' that on the right, 
so that yp^=za . P^xy on the left, and 
yrp z=z x'y' on the right. 

Divide the sjjan into sixteen equal 
parts 55^, 5^/, etc, and consider that the 
load, which is really uniformly dis- 
tributed, is applied to the arch at the 



pomts a, CKj, 



'1 5 



etc., in the verticals 



through &, 5j, ^/, etc.; so that the equal 
weights P are applied at each of the 
points on the left of a and the equal 
weights ^P dX each point on the right 
of a, while f i^ is applied at a. 

Take J as the pole of a force polj^gon 
for these weights, and lay off the weights 
which are applied at the left of a on the 
vertical through Z>g, viz., h^w^-=^\ 7^= the 
weight coming to a from the left ; 
w^ i^^=r:P=the weight applied at a, ; 
w^ ^€'3=P=:the weight applied at a^^ etc. 
Using h still as the pole, lay off h^ w^-= 
^P=the weight coming to a from the 
right; %ol w^^=-\ _P=:the weight applied 
at a/, etc. This amounts to the same 
thing as if all the weights were laid oft' 
in the same vertical. Part are put at 
the left and part at the right for con- 
venience of construction. Now draw 
hw^ until it intersects the vertical 1 at c^; 
then draw c c \\ hw ; and c^c^ \\ hw^y 



IN GRAPHICAL STATICS. 



15 



etc. In the same manner draw hw^ to 
c/ ; then c/ c/ || hw„'^ etc. Then the 
broken line hc^. . . c^ is the equilibrium 
polygon due to the weights on the left 
of a, and hc^' • . . c/ is that due to the 
weights on the right. Had the polygon 
been constructed for the uniformly dis- 
tributed load (not considered as concen- 
trated), on the left we should have a 
parabola passing through the points 
hc^ . . . Cg, and another parabola on the 
rio-ht through he' . . . cj . From the 



^ through hc^' . . . cj 
properties of this parabola it is easily 
seen that c^ must bisect u\w.^, as cj must 
also bisect w/ u\' ; which fact serves to 
test the accuracy of our construction. 
This test is not so simple in cases of 
more irregular loading. 

The equilibrium polygon c^ h c/ is that 
due to the applied weights, but if these 
weights act on a straight girder with 
fixed ends, this manner of support re- 
quires that the total bending be zero, 
when the sum is taken of the bending 
at the various points along the entire 
girder ; for, the position of the ends 
does not change under the action of the 
weights, hence the positive must cancel 
the negative bending. To expi*ess this 
by our equations : 

2/J=e. ^(J/)=rO .-. ^(J/)=:0. 

This is one of two conditions which 
are to enable us to fix the position of the 
true closing line h^h^ in this case. The 
other condition results from the fact 
that the algebraic sum of all the deflec- 
tions of this straight girder must be 
zero if the ends are fixed horizontally. 

This is evident from the fact that 
when one end of a girder is built in, if 
a tangent be drawn to its neutral axis 
at that end, the tangent is unmoved 
whatever deflections may be given to 
the girder; and if the other end be also 
fixed, its position with reference to this 
tangent is likewise unchanged by any 
deflections which may be given to the 
girder. To express this by our equations: 

]ld=f. ^ (J[f£C)=:0 .-. ^ {Mx)^^ 

The method of introducing these con- 
ditions is due to Mohr. Consider the 
area included between the straight line 
Cg c'/ and the polygon Cg h cj as some 
species of plus loading ; we wish to find I 
what minus loading will fulfill the above 
two conditions. Evidently the whole 



tive load c^ c 
in two parts. 

^8 < ' 



negative loading must be equal numeric- 
ally to the whole positive loading, if we 
are to have ^ {M)^=^0. Next, as the 
closing line is to be straight, the nega- 
Ag Ag' may be considered 
, viz., the two triangles, 
' Ag and c^' Ag Ag'. Let the whole 
span be trisected at t and t\ then the 
total negative loading may be considered 
to be applied in the verticals through 
t and t\ since the centers of gravity of 
the triangles fall in these verticals. 
Again, the positive loading we shall find 
it convenient to distribute in this man- 
ner : viz., the triangle c^ h cj applied in 
I the vertical through b, the parabolic area 
h c^ . . . c^ in the vertical 4 which con- 
tains its center of gravity, and the para- 
bolic area hc^' . . . cj in 4'. 

Now these areas must be reduced to 
equivalent triangles or rectangles, with 
a common base, in order that we may 
compare the loads they represent. Let 
the common base be half the span : then 
hh^-=^2jp' is the positive load due to the 
triangle c^ h c^' ; and § c^ c^-^pp^ and 
f c/ c^'=p'p^' are the positive loads due 
to the parabolic areas. 

Now assume any point ^ as a pole 
for the load line p^}^-! ^nd find the center 
of gravity of the positive loading by 
drawing the equilibrium polygon, whose 
sides are parallel to the lines of this 
force polygon : viz., use qp)^ and qji) as 
the 1st and 2nd sides, and make^g' || $'^9', 
and q'q^ \\ qp>^ . The first and last sides 
intersect at q^\ therefore the center of 
gravity of the positive loads must lie in 
the vertical through q^. 

Now the negative loading must have 
its center of gravity in the same vertical, 
in order that the condition ^'(J!^) = 
may be satisfied, for it is the numerator 
of the general expression for finding 
the center of gravity of the loading. 
The question then assumes this form : 
what negative loads must be applied in 
the verticals through t and t' that their 
sum may be p^p^^ and that they may 
have their center of gravity in the verti- 
cal through q^. 

The shortest way to obtain these two 
segments of p^p^ is to join ^^ and r' 
which are in the horizontals through 
p^ and ^/, and draw an horizontal 
through q^^ which is the intersection of 



r r' with the vertical through q^ ; then 
are the required segments 



rr^ and r' r^ 



16 



NEW CONSTEUCTIONS 



of the negative load. For, let Tr^=^p^'p^ 
^nd take r' as the pole of the load r r,^ ; 
then, since i\q^ \\ o\r' and g^o^' \\ r r' we 
have the equilibrium polygon i\ q^ r' ful- 
filling the required conditions. 

Now these two negative loads t^t^-=^ 
r'r' and rr^, are the required heights of 
the triangles c^ li^ c^ and Cg c/ h^ \ there- 
fore lay off Cg h^-=T' T^ and c^h^^^^rr^. 

The closing line h^ li^ can then be 
drawn, and the moments bending the 
straight girder will then be proportional 
to Aj Cj, h„ ^2, etc., the points of inflexion 
being where the closing line intersects 
the polygon. If the construction has 
been correctly made, the area above the 
closing line is equal to that below, a test 
easy to apply. 

Let us now turn to the consideration 
of the curve of the arch itself, and treat 
it as an equilibrium polygon. Since the 
rise of the arch is such a small fraction 
of the span, the curve itself is rather flat 
for our purposes, and we shall therefore 
multiply its ordinates ah^ a^ b^, etc., by 
any number convenient for our purpose: 
in this case, say, by 3. We thereby get 
a polygon d^dd^ such that d b^=^ ah, 
d^b^-—^d^b^, etc. If a curve be de- 
scribe* I through d^. . . d . . . d^' it will be 
the arc of an ellipse, of which d is the 
extremity of the major axis. 

If we wish to find the closing line k^Jc/ 
of this curve, such that it shall make 
:^ (Md)=:0 and :S {Mdx) = 0, the same 
process we have just used is here appli- 
cable ; but since the curve is symmetri- 
cal, the object can be effected more 
easily. By reason of the symmetry 
about the vertical through b, the center 
of gravity of the positive area above the 
horizontal through b lies in the vertical 
through b. The center of gravity of the 
negative area lies there also ; hence the 
negative area is symmetrical about the 
center vertical; the closing line must then 
be horizontal. It only remains then to find 
the height of a rectangle having the same 
area as the elliptical segment, and hav- 
ing the span for its base. This is done 
very approximately by taking (in this 
case where the span is divided into 16 
equal segments) J the sum of the ordi- 
nates b^ d^, etc. 

We thus find the height bk and the 
horizontal through k is the required 
closing line. 

Before effecting the comparison which 



we intend to make between the poly- 
gons c and d (as we may briefly desig- 
nate the polygons Cg b c/ and d^ d d^), let 
us notice the significance of certain oper- 
ations which are of use in the construc- 
tion before us. One of these is the 
multiplication of the ordinates of the 
polygon or curve a to obtain those of d. 
If a was inverted, certain weights might 
be hung at the points a^, a^, etc., such 
that the curve would be in stable equi- 
librium, even though there are flexible 
joints at these points. Equilibrium 
would still exist in the present upright 
position under these same applied 
weights, though it would be unstable. 
If now, radiating from any point, we 
draw lines, one parallel to each of the 
sides aa^, a^ a^, aa^\ etc., of the polygon, 
then any vertical line intersecting this 
pencil of radiating lines will be cut by it 
in segments, which represent the relative 
weights needed to make a their equilibri- 
um polygon. By drawing the vertical line 
at a proper distance from the pole, its 
total length, i. e., the total load on the 
arch can be made of any amount we 
please. The horizontal line from the 
pole to this vertical will be the actual 
horizontal thrust of the arch measured 
on the same scale as the load. If a like 
pencil of radiating lines be drawn paral- 
lel to the sides of the polygon d and the 
load be the same as that we had sup- 
posed upon the polygon a, it is at once 
seen that the pole distance for d is one- 
third of that for a ; for, every line in d 
has three times the rise of the corre- 
sponding one in «, and hence with the 
same rise, only one-third the horizontal 
span. The increase of ordinates, then, 
means a decrease of pole distance in the 
same ratio, and vice versa. As is well 
known, the product of the pole distance 
by the ordinate of the equilibrium poly- 
gon is the bending moment. This pro- 
duct is not changed by changing the 
pole distance. 

Again, suppose the vertical load-line 
of a force polygon to remain in a given 
position, and the pole to be moved ver- 
tically to a new position. No vertical 
or horizontal dimension of the force 
polygon is affected by this change,, 
neither will any such dimension of the 
equilibrium polygon corresponding to 
the new position of the pole be differ- 
ent from that in the polygon corre- 



IN GRAPHICAL STATICS. 



17 



spending to the first position of the pole ; 
the direction of the closing line, how- 
ever, is changed. Thus we see that the 
closing line of any equilibrium polygon 
can be made to coincide with any line 
not vertical, and that its ordinates will 
be unchanged by the operation. It is 
unnecessary to draw the force polygon 
to effect this change. 

Now to make clear the relationship 
between the polygons c and d^ let us 
suppose, for the instant, that the poly- 
gon e has been drawn by some means 
as yet unknown, so that its ordinates 
from c?, viz., e^d^=^y^, e„d^^=y^, etc., are 
proportional to the actual moments 3fe 
which tend to bend the arch. 

The conditions which then hold re- 
specting these moments Me, are three : — 

:E(Me) = 0, 2{Mex)^0, :S {Meij) = 0. 

The first condition exists because the 
total bending from end to end is zero 
when the ends are fixed. The second 
and third are true, because the total de- 
flection is zero both vertically and hori- 
zontally, since the span is un variable as 
well as the position of the tangents at 
the ends. These results are m accord- 
ance with Prop. V. IsTow by Prop. Ill 
these moments Me are the differences of 
the moments of a straight girder and of 
the arch itself ; hence the polygon e is 
simply the polygon c in a new position 
and with a new pole distance. As 
moments are unchanged by such trans- 
formations, let us denote these moments 
by Mc We have before seen that 

2(Mc) = 0, and :S (Mcx)^0 
Subtract 

.-. :S{Mc-Me)=0, and 2{Mc-Me)x=0 
.'. :2{Md) — and 2 {Mdx) = 

From this it is seen that the polygon 
d must have its closing line fulfill the 
same conditions as the polygon c. This 
is in accordance with Prop. IV. 

Again, :E (Mey) — :E {Mc — Md) y=0 

.-. 2{Mcy) = 2{3Idy). 

This last condition we shall use for 



determining the pole distance of tlie 
polygon e, which is one-third of the 
actual thrust of the arch measured on 
the scale of the weights w^ to^, etc. The 
physical significance of this condition 
may be stated according to Prop. V, 
thus : if the moments Md are applied to 
a uniform vertical girder bd at the points 
b, b"^ 6/', b" , etc., at the same height 
with &g, d^, etc., they will cause the same 
total deflection xd^=- e .'2. [Mdy) as wdll 
the moments Mc when applied at the 
same points. Hence if Md are used as 
a species of loading, we can obtain the 
deflection by an equilibrium polygon. 
Suppose the load at d^ is d^ /<;„ and that 
at d^ is c?g /(^g, etc., then that at b^ is 
^ ^g k^. This approximation is sufficiently 
accurate for our purposes. 

N'ow lay off on l^ I J as a load line 
dm^=ib^k^, 7n^m^ = d^k^, m^m^^d^h^^ 
etc. The direction of these loads must 
be changed when they fall on the other 
side of the line k\ e.g., 711. m^^i^h, d,. 
Ir this process be continued through the 
entire arch m/ (not drawn) will fall a& 
far to the right of d as m^ does to the 
left, and the last load will just reach 
to d again. This is a test of the cor- 
rectness with which the position of the 
line k^ k/ has been found. Now using 
any point as b for a pole, draw bm^ to /\y 
then draw //^ || ^m^, /;/; || bm^, etc. 
The curve bf is then the exaggerated 
shape of a vertical girder bd, fixed at b^ 
under the action of that part of moments 
Md which are in the left half of the 
arch. The moments Md on the right 
may act on another equal girder, havings 
the same initial position bd, and it will 
then be equally deflected to the right of 
bd. This is not drawn. 

Again, suppose these vertical girders 
fixed at b are bent instead by the 
moments Mc. We do not know just 
how much these moments are, though we 
do know that they are proportional to 
the ordinates of the polygon c. There- 
fore make dn^ — i- A„ c„, n„ n„ = h„ c 



n„ 



2 8 8? 8 7 '"7 ^'i'i 

n^ = h^ c„ etc. When all these loads 



are laid off, the last one n/ d=^^hj c/ 
must just return to d. This tests the- 
accuracy of the work in determining the 
position of h^ lij. 

Now using ^ as a pole as before, con- 
struct the deflection curves bg and bg'. 
Since these two deflections, viz., 2 dj^ 
and gg' ought to be the same, this fact 



18 



NEW CONSTRUCTIOIS'S 



informs lis that each of the ordinates 
Aj Cj, A2 c^, must be increased in the ratio 
of ^ gg' to f^/", in order that when they 
are considered as loads, they may pro- 
duce a total deflection equal to 2 df. 
To etfect this, lay off bj=^clf and bi= 
i gg\ and draw the horizontals through 
i and /. At any convenient distance 
draw the vertical i^j^^ and draw bi^ and 
hj^. These last two lines enable us to 
elfect the required proportions for any 
ordinates on the left, and these or two 
lines of the same slope on the right to 
do the same thing on the right. M. g. 
lay off the ordinate bij=^k^' c/, then 
the required new ordinate is bj^' . Then 
lay off k^' e^'=- bj\'. In the same man- 
ner find k e from A b, and Jc^ e^ from Ag c^. 
In the same manner can the other ordi- 
nates k^ e^, etc., be found ; but this is 
not the best way to determine the rest 
of them, for we can now find the pole 
and pole distance of the polygon e. 

As we have previously seen, the pole 
distance is decreased in the same ratio 
as the ordinates of the moment curve 
are increased, therefore prolong bi^ to v^, 
and draw a horizontal line through v^ 
intersecting bj\ at v^ and the middle ver- 
tical at v^ ; then is v^ "^0 the pole dis- 
tance decreased in the required ratio. 
Hence we move up the weight-line to^ w^ 
to the position u^ u^ verticall}^ through 



<v^\ and for convenience, lay off the 
weights 10^' w^' at w/ 2^/, etc. 

Furthermore, we know that the new 
<?losing-line is horizontal. To find the 
position of the pole so that this shall 
occur, draw bv parallel to A/ig, and from 

V the horizontal vo. As is well known, 

V divides the total weight into the two seg- 
ments, which are the vertical resistances 
of the abutments, and if the pole o is 
on the same horizontal with v, the 
closing line will be horizontal. 

Now having determined the positions 
■of the points 6g, e, e/, starting from one 
of them, say 6g, draw gg e„ || oic^^ e^^e^ \\ ou^^ 
etc.; then if the work be accurate, the 
polygon will pass through the other two 
points e and e^ . The bending moments 
of the arch d or the arch a at «j, a^, etc., 
is the product of the pole distance 
'o^xi^-=^X)'o by the ordinates d^e^^ ^^2^25 
etc., respectively, and between these 
points a similar product gives the mo- 
anent with sufiicient accuracy. It would 
ibe useful for the sake of accuracy to 



multiply the ordinates of the arch by 
some number greater than 3. 

As a final test of the accuracy of the 
work, let us see whether ^ {Mey) is ac- 
tually zero, as should be. At cZ„ for ex- 
ample, y=idj^.) and 3Ie is i:>roportional 
to d^ e,. Then d^ s^ is proportional to 
3Iey at that point if e^ s^ is the arc of 
a circle, of which e. l^ is the diameter. 
Similarly find dj s' etc. When e, for 
example falls above c?^, the circle must 
be described on the sum of L c?, and d, e, 

as a diameter, and d^ s^ is proportional 
to a moment of different sign from that 
at d.. We have distinguished the sign 
of the moments at the different points 
along the arch, by putting different 
signs before the letter s. It would have 
been slightly more accurate to have used 
only one-half the ordinates ^g e^ and 
b^' gg', but as they nearly equal in this 
case and of opposite sign, we have in- 
troduced no appreciable error, 

Now at any point s lay off ss^ — d^s^, 
and at right angles to it 5^ ^g = ^8 ^s? then 
at right angles to the hypothenuse ss^ 
make s^s'=^d's' etc. Then the sum 

8 5 5 5" 

of the positive squares is ss/, and simi- 
larly the sum of the negative squares is 
ss^. If these are equal, then ^ {3Iey) 
vanishes as it should, and the construc- 
tion is correctly made. 

It would have been equally correct to 
suppose the. two vertical girders fixed at 
c?, and bent by the moments acting. We 
could have determined the required ratio 
equally well from this construction. 
Further, in proving the correctness of 
the construction by taking the algebraic 
sum of the squares, we could have reck- 
oned the ordinates, y, from any other 
horizontal line as well as from l^ IJ. 

To find the resultant stress in 
the different portions of the arch, 
we must prolong v'o to o\ say, 
(not drawn) so that the pole distance 
v^o'=^S v'o ; then if we join o' and u^, 
o'u^ will be the resultant stress in the 
segment b^a^', o'u^ will be the stress in 
<2. ttg, etc., measured in the same scale as 
the weights to^ lo^, etc. This resultant 
stress is not directly along the neutral 
j axis of the arch. 

I The vertical shearing stress is construct- 
j ed in the same manner as for a girder, 
j by drawing one horizontal through w^ 
I between the verticals 7 and 8, another 



IN GRAPHICAL STATICS. 



19 



through w^ between 7 and 6, etc. (not 
drawn). Then the shear will be the ver- 
tical distance between vo and these hori- 
zontals through w^^ w„ etc. It is seen 
that the shear will change sign on the 
vertical through b^ with our present 
loading. 

The actual position of the vertical 
through the center of gravity of the 
load may be found by prolonging the 
first and last sides of the polygon c. A 
weight = k jP =^ 10^ 10^ ought, however, 
first to be applied at h„, and another 
■= \ P=: ii\' i(\' at hj. The shearing 
stress under a distributed load will 
actually change sign on the vertical so 
found. It will not pass far however 
fi'om by 

Tlie resultant stress is the resultant of 
the horizontal thrust and the vertical 
shearing stress, and it can be resolved 
into a tano-ential thrust alono; the arch 
and a normal shearing stress. This 
resolution will be effected in Fig. 3 of 
the next chapter. 

As to the position of the moving load 
which will produce the maximum bend- 
ing moments, we may say that the posi- 
tion chosen, in which the moving load 
covers one-half the span, gives in general 
nearly this case. It is possible, how- 
ever, to increase one or two of the 
moments slightly by covering a little 
more than half the span with the mov- 
ing load. 

The loading which produces maximum 
moments will be treated more fully in 
subsequent chapters. 

The maximum resultant stress and 
maximum vertical shear occur in gen- 
eral when the moving load covers the 
whole span. The construction in this 
case is much simplified, as the poly- 
gon G is then the same on the right of 
b as it now is on the left, and the 
center of gravity of the area is in the 
center vertical ; so that the closing line 
h^ h/ is horizontal, and can be drawn 
with the same ease as k^ k/ was drawn. 
We shall not, even in this case, be under 
the necessity of drawing the curves b(/ 
and bg\ which would be both alike ; for, 
as may be readily seen, the sum of the 
positive moments Mc on the left must 
be very approximately equal to the 
positive moments 3fd on the left, and 
the same thing is true for the negative 



moments at the left. The same two 
equalities hold also on the right. From 
this we at once obtain the ratio by which 
the ordinates of the polygon e must 
be altered to obtain those of the poly- 
gon e. 

This last approximation also shows us 
that for a total uniform load, the four 
points of inflection when the bending 
moment is zero, lie two above and two 
below the closing line. It is frequently 
a sufficiently close approximation in the 
case when the moving load covers only 
part of the span to derive the ratio 
needed by supposing that the sum of all 
the ordinates, both right and left, above 
the closing line in the polygon c must 
be increased, so that it shall equal the 
corresponding sum in the polygon d. 
If the sums taken below the closing 
lines give a slightly different result, take 
the mean value. 

Thus the single construction we have 
given in Fig. 2, and one other much 
simpler than this, which can be ob- 
tained by adding a few lines to 
Fig. 2, give a pretty complete deter- 
mination of the maximum stresses on 
the assumptions made at the commence- 
ment of the article. 

One of these assumptions, viz., that 
of constant cross section {i. e. jT^ con- 
stant), deserves a single remark. In 
the St. Louis Arch J was increased 
one-half at each end for a distance of 
one-twelfth of the span. This very 
considerable change in the value of J 
slightly reduced the maximum moments 
computed for a constant cross section. 
From other elaborate calculations, par- 
ticularly those of Heppel,* on the Britan- 
nia Tubular Bridge, it appears that the 
variation in the moments caused by the 
changes in cross section, which will 
adapt the rib to the stresses it must sus- 
tain, are relatively small, and in ordinary 
cases are less than five per cent, of the 
total stress. The same considerations 
are not applicable near the free ends of 
a continuous girder, where J may theo- 
retically vanish. In the case before us, 
where the principal part of the stress 
arises not from the bending moments, 
but from the compression along the 
arch, the effect of the variation of 7" is 
very inconsiderable indeed. 

* Philosophical Magazine, Vol. 40, 1870. 



20 



NEW CONSTEUCTIONS 



CHAPTER III. 

ARCH EIB WITH FIXED ENDS AND HINGE 
JOINT AT THE CROWN. 

Let the curve a of Fig. 3 represent 
the proportions of the arch we shall use 
to illustrate the method to be applied to 
arches of this character. The arch a is 
segmental in shape, and has a rise of one- 
fifth of the span. It is unnecessary to 
assume the particular dimensions in feet, 
as the above ratio is sufficient to deter- 
mine the shape of the arch. 

The arch is supposed to be fixed in the 
abutments, in such a manner that the 
position of a line drawn tangent to the 
curve a at either abutment is not changed 
in direction by any deflection which the 
arch may undergo. At the crown, how- 
ever, is a joint, which is perfectly free to 
turn, and which will, then, not allow the 
propagation of any bending moment 
from one side to the other. In order 
that we may effect the construction more 
accurately, let us multiply the ordinates 
of the curve a by some convenient num- 
ber, say 2, though a still larger multi- 
plier would conduce to greater accuracy: 
We thus obtain the polygon d. 

Having divided the span h into twelve 
equal parts h^ h^., etc, (a larger number of 
parts would be better for the discussion of 
an actual case), we lay off below the hori- 
zontal line b on the end verticals, lengths 
which express on some assumed scale the 
weights which may be supposed to be 
concentrated at the points of division of 
the arch. If a I is the depth of the load- 
ing on the left and al'—-\al that on the 
right, then h^iv^ + h^iu^^^ the weight con- 
centrated at a\ w^w^-= the weight at a^\ 
w^ w./= the weight at a/, etc. Using 
6 as a pole, draw the equilibrium polygon 
c, whose extremities Cg and c/ bisect 
w^ 10^ and w^ i/;/ respectively. 

Now to find the closing line of this 
equilibrium polygon so that its ordinates 
shall be proportional to the bending mo- 
ments of a straight girder of the same 
span, and of a uniform moment of inertia 
Z, which is built in horizontally at the 
ends and has a hinge joint at its center; 
we notice in the first place that the bend- 
ing moment at the hinge is zero, and 
hence the ordinate of the equilibrium 
polygon at this point vanishes. The 
closing line then passes through h the 
point in question. Furthermore it is 



evident that if we consider the parts of 
the girder at the right and left of the 
center as two separate girders whose 
ends are joined at the center, these ends 
have each the same deflection, by reason 
of this connection. 

This is expressed by means of our 
equations by saying that ^{Mx) when 
the summation is extended from one end 
to the center is equal to 2{Mx) when the 
summation is extended from the other 
end to the center, for these are then pro- 
portional to the respective deflections of 
the center. We may then write it thus : 

4^ (Mx) = 4^iMx) 

The equation has this meaning, viz : 
that the center of gravity of the right 
and left moment areas taken together is 
in the center vertical : for, taking each 
moment Jif as a weight, x is its arm, and 
Mx its moment about the center. 

In order to find in what direction to 
draw the closing line through b so that 
it shall cause the moment areas together 
to have their center of gravity in fhe 
center vertical through b, let us draw a 
second equilibrium polygon using the 
moment areas as a species of loading. 

The area on the left included between 
any assumed closing line as bb^ (or bh^) 
and the polygon bc^ may be considered 
to consist of a positive triangular area 
bcj)^ (or bcji^) and a negative parabolic 
area bc^c^c^; and similarly on the right a 
positive area bcjb^' (or bcjhj) and a nega- 
tive area bc^'cjc^'. 

At any convenient equal distances from 
the center as at p and p', lay off these 
loads to some convenient scale. It is, 
perhaps, most convenient to reduce the 
moment areas to equivalent triangles 
having each a base equal to half the 
span: then take the altitudes of the tri- 
angles as the loads. This we have done, 
so that pp=^c^c^,^ and p'p'—^G^c^. 
Now assume, for the instant, that closing 
line is bp^^ which of course is incorrect, 
and make p,p^ — b^c^ and p'p^^b^c^., 
then these are the loads due to the posi- 
tive triangular areas at the left and right 
respectively, while pp)^ and p'p^ are the 
negative parabolic loads. 

Take o' as the pole of these loads, then 

pjp' may be taken for the first side of the 

second equilibrium polygon. Draw pq 

II o'pj^ and p'q' \\ o'p^\ and then from q 



IN GRAPHICAL STATICS. 



21 



and q' draw parallels to o'p^' respective- 
ly. These last sides intersect at q„. The 
vertical through q.^ then contains the 
center of gravity of the moment areas 
when h^ b^ is assumed as the closing 
line. 

A few trials will enable us to find the 
position of the closing line which causes 
the center of gravity to fall on the center 
vertical. We are able to conduct these 
trials so as to lead at once to the required 
closing line as follows. Since, evidently, 

of the positive 



sum 



Cg', it is seen that the 
loads is constant. 



Therefore niake^:>2P3=/5//>3' and m^q 2'^iP2 
and 2^^p2 ^s the positive loads, in the 
same manner as we used p>\Pi ^"^ P' iP' i 
previously. 

This will be equivalent to assuming a 
new position of the closing line. The 
only change in the second equilibrium 
polygon will be in the position of the 
last t vvu sides. These must now be drawn 
parallel to o'}')^ and o'p^' respectively; 
and they intersect at q^. The vertical 
through ^„ contains the center of gravity 
for this assumed closing line. Another 
trial gives us q^. 

Xow if the direction of the closing 
line had changed gradually, then the in- 
tersection of the last sides of the second 
equilibrium polygon would have de- 
scribed a curve through q^^ q^ and q^. If 
one of these points, as q^^ is near the cen- 
ter vertical, then the arc of a circle q.^q^ 
q^, will intersect it at 5. indefinitely near 
to the point where the true locus of the 
points of intersection would intersect the 
center vertical. 

Let us assume that g. is then deter- 
mined wdth sufficient exactness by the 
circular arc q^q/i^^^ and draw qq.^ and q'q^ 
as the last two sides of the second equili- 
brium polygon. Now draw o'p^ \\ qq^ 
and o'p^ II §''5'., then/)j_2>5=<^6^^6 and^:?/^/ 
=:c/A/ are the required positive loads, 
and liph^ is the position of the closing 
line such that the center of gravity of 
the moment areas is in the center verti- 
cal. 

It is evident that the closing line of the 
polygon d considered as itself an equilib- 
rium polygon is the horizontal line 
through c?, for that will cause the center 
of gravity of the moment areas on the 
left and right, between it and the polygon 
c?, to fall on the center vertical. 

The next step in the construction is to 



apply Prop. IV, for the determination of 
the bending moments. 

That Prop. IV is true for an arch of 
this kind is evident; for, the loading 
causes bending moments proportional to 
the ordinates A^c^, AgC^, etc., while the arch 
itself is fitted to neutralize, in virtue of 
its shape, moments which are proportional 
to /v/?2? ^'3*^^35 6tc. The differences of 
the moments represented by these ordi- 
nates are what actually produce bending 
in the arch. 

Xow the ordinates of the type lie are 
not drawn to the same scale as those of the 
type hd^ for each was assumed regardless 
of the other. In order that we may find 
the ratio in which the ordinates he must 
be changed to lay them off on the same 
scale as hd it is necessary to use another 
equation of condition imposed by the 
nature of the joint and supports, viz: 

or tl^ {Mi~M,)y = 4^,{Ma-M,)v 

. The left hand side of the equation is the 
horizontal displacement {i.e., the total 
, deflection) of the extremity a of the left 
! half of the arch, due to the actual bend- 
ing moments {Md — Mc) acting upon it: 
and the rio-ht hand side is the horizontal 
displacement of a the extremity of the 
right half of the arch due to the moments 
i actually bending it. These are equal be- 
' cause connected by the joint. 
I The construction of the deflection 
I curves due to these moments will enable 
I us to find the desired ratio. 

The ordinates kd and he are rather 
longer than can be used conveniently, to 
I represent the intensity of the moments 
i concentrated at d^,d^y etc, and c^,c^, etc.: 
: so we will use the halves of these quan- 
I titles instead. Therefore lay off dm^-=- 

and also dn^ = ^ h^c^, 7i^n^ =: J h^c., etc. 
I We use only one-quarter of each end 
' ordinate because the moment area sup- 
posed to be concentrated at each end has 
i only one half the width of the moment 
! areas concentrated at the remaining 
I points of division. 

j Using 5 as a pole we find the deflection 
' curve fb due to the moment 3fa or Jla 
; and the deflection curve gb due to the 
, moments 3fc on the left. On the right 



22 



NEW CONSTRUCTIONS 



we should find a deflection df'^df not 
drawn, and similarly a deflection dg' not 
equal to dg. 

Now the equation we are using requires 
that the ordinates he shall be elongated 
so that when used as weights the deflec- 
tions shall be identical : i.e., we must 
have df^=^^gg'. To effect the elongation, 
lay off aj:=df and ai=^gg'; and at any 
convenient distance on the horizontals ii^ 
and j'j\ draw the vertical i^j\; then the 
lines ai^ and aj\ will effect the required 
elongation. For example, lay off a^g=. 
AgCg, from which we obtain aj\=k^e^ for 
the left end ordinate, and similarly aj/= 

The pole distance tt^^ of the original 
polygon c must be shortened in the 
same ratio in which the ordinates are 
elongated. Hence the new pole distance 
of the polygon e is tt.^. 

Since ^'g^/ is the closing line of the 
polygon e, and is horizontal, the pole of 
e is o, on the horizontal through h^\ for, 
AgiOg is the part of the applied weight 
sustained by the left support. 

Now if the weight line be moved up 
to t^ so that the applied weights are u^u^ 
at the center, etc., and o is the pole, the 
polygon e may be described starting from 
d, and it will finally cut off the end ordi- 
nates ^g6g and ^g^^g^ before obtained. 
Then will the ordinates of the type de 
be proportional to the moments actually 
bending the arch, and the moments will 
be equal to the products of de by tt^.^ in 
which de is measured on the scale of 
distance, and tt^ on the scale adopted for 
the weights w^io^., etc. 

The accuracy of the construction is 
finally tested by taking ^(<^5)'=0, an 
equation deduced from I" {Ma— Mc)y=^0, 
as explained in the previous article upon 
the St. Louis Arch. It is unnecessary to 
explain the details of this construction 
since as appears from Fig. 3 it is in all 
respects like that in Fig. 2. 

Now let us find the intensity of the 
tangential compression along the arch 
and of the shearing normal to the arch. 
Since the pole distance tt^ refers to the 
difference of ordinates between the poly- 
gons d and 6, whose ordinates are double 
the actual ordinates, if we wish now to 
return to the actual arch a whose ordi- 
nates are halves of the ordinates of d, 
we must take a pole distance U^ = 2tt^ and 
move the weight line so that it is the 



vertical through t^. Then U^ is the actual 
horizontal thrust of this arch due to the 
weights; and ov^ is the resultant stress 
in the segment a^b^ of the arch, which 
may be resolved into two components 
org and v^9\ respectively parallel and per- 
pendicular to a^b^. 

Then are or^ and ^g^g respectively, the 
thrust directly along, and the shear di- 
rectly across the segment aj>^ of the 
arch. Similarly or^ and v^r^ represent 
the thrust along, and the shear across 
the segment a^a^, and so on for other 
segments. These quantities are all 
measured in the same scale as that of the 
applied weights. 

The shear changes sign twice, as will 
be seen from inspection of the directions 
in which the quantities of the type vr 
are drawn. The shear is zero wherever 
the curves d and e are parallel to each 
other. Thus the shear is nearly zero at 
Z>g, at a^ and at some point between a/ 
and a/. 

The maxima and minima shearing 
stresses are to be found where the incli- 
nation between the tangents to the curves 
d and e are greatest. 

The statements made in the previous 
article, respecting the position of the 
moving load which causes maximum 
bending moments, are applicable to this 
kind of arch also. 

The maximum normal shearing stress 
will occur for the parts of the arch near 
the center, when the moving load is near 
its present position, covering one half of 
the arch. But the maximum normal 
shearing stress near the ends, may occur 
when the arch is entirely covered by the 
moving load, or when it may occur when 
the moving load is near its present posi- 
tion, it being dependent upon the rise of 
the arch, and the ratio between the mov- 
ing and permanent load. 

The maximum tangential compressions 
occur when the moving load covers the 
entire arch. The stresses obtained by 
the foregoing constructions, go upon the 
supposition that the arch has a constant 
cross-section, so that its moment of iner- 
tia does not vary, and no account is 
taken of the stresses caused by any 
changes of the length of the arch rib, 
due to variations of temperature or other 
causes. These latter stresses we shall 
now investigate for both of the kinds of 
arches which have been treated. 



IN GRAPHICAL STATICS. 



23 



CHAPTER TV. 

TEMPEKATUEE STRAINS. 

It is convenient to classify all strains 
and stresses arising from a variation in 
the length of the arch, under the head 
of temperature, as such stresses could 
evidently have been brought about by 
suitable variations of temperature. 

The stresses of this kind which are of 
sufficient magnitude to be worthy of con- 
sideration, besides temperature stresses 
are of two kinds, viz. the elastic short- 
ening of the arch under the compression 
to which it is subjected, and the yielding 
of the abutments, under the horizontal 
thrust applied to them by the arch. 
This latter may be elastic or otherwise. 
It was, I believe, neglected in the com- 
putation of the St. Louis Arch, and no 
doubt with sufficient reason, as the other 
stresses of this kind were estimated with 
a sufficient margin to cover this also. 
Anything which makes the true span of 
the arch differ from its actual span 
causes strains of this character. By true 
span is meant the span which the arch 
would have if laid flat on its side on a 
plane surface in such a position that 
there are no bending moments at any 
point of it, while the actual span is the 
distance between the piers when the 
arch is in position. If the arch be built 
in position, but joined at the wrong tem- 
perature the true and actual spans do 
not agree and excessive temperature 
strains are caused. 

Taking the coefficient of expansion of 
steel as ordinarily given, a change of 
±80°F. from the mean temperature 
would cause the St. Louis Arch to be 
fitted to a span of about 3 J inches, greater 
or less than at the mean. 

The problem we wish to solve then is 
very approximately this : What hori- 
zontal thrust must be applied to increase 
or decrease the span of this arch by 3j 
inches, and what other stresses are in- 
duced by this thrust. In Fig. 4 the half 
span is represented on the same scale as 
in Fig. 2. The only forces applied to 
the half arch are an unknown horizontal 
thrust ^at ^g and an equal opposite 
thrust ^at a. The arch is in the same 
condition as it would be if Fig. 4 repre- 
sented half of a gothic arch of a span = 
2a5, of which a was one abutment, and h^ 
was the new crown at which a weio^ht of 



2^ was applied. The gothic arch would 
be continuous at the crown, but the 
abutment a would be mounted on rollers, 
so that although the direction of a tan- 
gent at a could not be changed, neverthe- 
less the abutment could afford no resist- 
ance to keep the ends from moving 
apart, i.e. there is no thrust in the direc- 
tion of «^, any more than there is along 
an ordinary straight girder. 

In order to facilitate the accurate con- 
struction, let us multiply the ordinates 
of a by 3 and use the polygon d instead. 
Now the real equilibrium polygon of the 
applied forces H^ is the straight line hlc^. 
By real equilibrium polygon is meant, 
that one which has for its pole distance, 
the actual thrust of the arch. As we 
are now considering this arch, H is the 
applied force, and the thrust spoken of 
is at right angles to S, We have just 
shown this thrust to be zero. We have 
then to construct an equilibrium polygon 
for the applied force II with a pole dis- 
tance of zero. The polygon is infinitely 
deep in the direction of H., and hence is 
a line parallel to H. This fixes its direc- 
tion. 

Its position is fixed from the considera- 
tion that the total bending is zero, (be- 
cause the direction of the tangents at 
the extremities a and h^ are unchanged), 
which is expressed by the equation 

This gives us the same closing line 
through h which we found in Fig. 2, and 
the ordinates of the type lid., are propor- 
tional to the moments caused by the 
horizontal thrust H. 

Now lay off dm^=i^JcJ)^, on^m^^k^d^y 
etc., as in Fig. 2. 

The problem of finally determining^ 
will be solved in two steps: 

1°. We shall find the actual values of 
the moments to which the ordinates kd 
are proportional; 

2°. We shall find ^by dividing either 
of these moments by its arm. 

By considering the equation 

lJyEI=2(My) 

given in Chapter I, in which Dy is 
the horizontal displacement, it is seen 
that if the actual moments are used for 
weights, and UI for the pole distance, we 
shall obtain, as the second equilibrium 
polygon, a deflection curve whose ordi- 



24 



NEW CONSTRUCTIOIS^S 



nates are the actual deflections due to 
the moments. By actual moments, actual 
deflections, etc, is meant, that all of the 
quantities in the equation are laid off to 
the scale of distance, say one 7i^^^ of the 
actual size. 

Now let the equation be written 

nBy . \EI=I{My). 

it 

From which it is seen that if the ordi- 
nates be multiplied by w, so that on the 
paper they are of the same size as in the 
arch, we must use one n^^^ of the former 
pole distance, all else remaining un- 
changed. 

Novy for the St. Louis Arch, jET= 
39680000 foot tons. Let us take 100- 
tons to the inch, as the scale of force : 
and since ^^=3 inches, the scale of dis- 
tance n is found from the proportion 

3 in. : : 51.8 ft. : : 1 \n = 210 nearly, 
and EI ~ 100 n^ — ^ in. nearly, 

which is the pole distance necessary to use 
with the actual deflection ^ of 3:J^in.= 
If in., in order that the moments may be 
measured to scale. As it is inconvenient 
to use so large a distance as 9 in. on our 
paper, let us take f of 9 in. =3j in. 
nearly :=^dz for the pole distance, and 
f of if in. =4§ in. =<:?2/, for the deflec- 
tion. 

Now with 2 as a pole and the weights 
dm^^ ni^m^., etc, draw the deflection curve 
bf, having the deflection =^df. The mo- 
ments Mci must be increased in such a 
ratio that the deflection will be increased 
from c7/ to dy. Therefore draw the 
straight lines hf and by, which will ena- 
ble us to effect the increase in the required 
ratio. For example, the moment dm^-=^bi 
is increased to hj^ and dm^-=.bj is increased 
to l)j^. Now measuring bj in inches and 
multiplying by 210 and by 100, we have 
found that 21000 fy"=1809 foot tons=the 
moment at d or a. 

And again, 21000 bj\=:SH1 foot tons 
=the moment at 5^. 

By measurement 210 dk=l7 ft. and 
210 ^>^=34.8 ft. 

.-. ir= 1809 -^-17 = 106 tons, -f 

or ^=:3747-v-34.8 = 108 tons — . 

These results should be identical, and 
the difference between them of less than 
2 per cent, is due to the error occasioned 



by using the polygon d instead of the 
curve of the ellipse, and to small errors 
in measurement. With a larger figure 
and the subdivision of the span into a 
greater number of parts this error could 
be reduced. The value of H found for 
the St. Louis Arch by computation was 
] 04 tons, but that was not on the suppo- 
sition of a uniform moment of inertia 7", 
and should be less than the value w^e 
have obtained. 

Now this horizontal thrust IT due to 
temperature and to any other thrusts 
of like nature as compression, etc, is of 
the nature of a correction to the thrust 
due to the applied weights. Thus in 
Fig. 2 we found Son' to be the thrust due 
to the applied weights, and on applying 
the correction we must use the two 
thrusts 3o?;^-f^and 3o^;'— ^as pole dis- 
tances to obtain equilibrium polygons 
whose ordinates reckoned from the arch 
a will, when multiplied by its pole dis- 
tance, give the true bending moments. 
The tangential and normal stresses can 
then be determined by resolution, pre- 
cisely as in Fig. 3. 

If it, however, appears desirable to 
compute separately the strains due to 
IT, this may be more readily done than 
in combination with the stresses already 
obtained. We have already seen suffi- 
ciently how the bending moments due 
to JTare found. Li fact the moments 
are such as would be produced by apply- 
ing ^at the point where the horizontal 
through k cuts the polygon d, for this is 
the point of no moment, and may be 
considered for the instant as a free end 
of each segment, to each of which J?" is 
applied causing the moments due to its 
arm and intensity. 

To find the tangential stress and shear, 
lay off in Fig. 4 av = II'aTid on it as a di- 
ameter describe a semicircle, and draw 
ar^ II a^a^, ar^ \\ ap^ etc.; then will ai\ be 
the component of ^along a^a^^ and vr^ be 
the component of H directly across the 
same segment. In a similar manner the 
quantities of which ar on the type are 
the tangential stresses and the quantities 
vr are the shearing stresses caused by 

The scale used for this last construc- 
tion is about fifty tons to the inch. 

Now jETis positive or negative accord- 
ing as the temperature is increased 
above or diminished below the mean, 



IlSr GRAPHICAL STATICS. 



23 



and these tangential and normal com- 
ponents, of course, change sign with 11. 

It should also be noticed in this connec- 
tion that thrusts and bending moments, 
which are numerically equal but of op- 
posite sign, are induced by equal con- 
tractions and expansions. 

The stresses due to variation of tem- 
perature in the arch of Fig. 3, having a 
center joint, are constructed in Fig. 5. 

It is evident from reasoning similar to 
that employed for the case just discussed, 
that the closing line dl:^ of the polygon 
d is the equilibrium polygon of the thrust 
^induced by variation of temperature. 
Suj^pose we have changed the equation 
of deflections to the form. 



mDo 



EI 



tnn 






in which, if niDy^zdy and El-^mn^^zdz^ 
then the moments M and the ordinates 
y will be laid off on the scale of 1 to n. 
This is equivalent to doing what was 
done in the previous case, where m was 
equal to f. The remainder of the pro- 
cess is that previously employed. 

It should be noticed that we have in 
Figs. 4 and 5, incidentally discussed two 
new forms of arches, viz: in Fig. 4 that 
of an arch having its ends fixed in direc- 
tion, but not in position; i.e., its ends 
may slide but not turn, and in Fig. 5, 
that of an arch sliding freely and turn- 
ing freely at the ends. The first of these 
arches has the same bending moments as 
a straight girder, fixed in direction at the 
ends, and the second of them has the same 
bending moments as a simple girder sup- 
ported at its ends. 

Errata. — The measurements of Fig. 4 
given on page 24 do not agree with the 
scale on which the drawing is engraved. 
The following equations and quantities 
agree with the dimensions of Fig. 4, and 
are to be substituted instead of those 
given on page 24. 

Let the scale of force be 100 tons to 
the inch, and since 5c?=4^ inches, 4 J in. 
: 51.8 ft. : : 1 : n=:140 nearly, and EI^ 
100n^ = 20 in. nearly, which is the pole 
distance to use with the actual deflection 
of the half span = lf in. 

Now take one fourth of this pole dis- 
tance = 5 in. = dz^ and four times the 
deflection = 6 J in. = dy, as being more 
convenient to use; the moments, which 



are the products of the deflections by 
the pole distance, will be unchanged by 
this process. 

Now increase the ordinates in such a 
ratio that the deflection will be increased 
from df to dy. For example, the mo- 
ment dm^=-hi is increased to hj, and dm^ 
= bi^ is increased to bj^. Now by meas- 
uring b] in inches and multiplying by 
140 and by 100 we have found 14000 bj = 
1809 foot tons=the moment at a or d. 
And again, 14000 bj\=374:1 foot tons = 
the moment at b^. 

By measurement, 140 dk=11 it. 

and 140 ^»^^=34.8 ft. 

.-. ^=1809-^-l7 = 106 tons -f, 
or ^=3747-r-34.8 = 108 tons — . 

Near the bottom of the second column 
of page 24, instead of ar^, ar^^ vr^, ar, vr, 
read av^. av^, vv^, av, vv. 

The scale used in the last construction 
in Fig. 4, is about 33 J tons to the inch. 

UNSYMMETRICAL AECHES. 

The constructions which have been 
given have been simplified somewhat 
by the symmetry of the right and left 
hand halves of the arch, but the meth- 
ods which have been used are equally 
applicable if such symmetry does not 
exist, as it does not, if, for example, the 
abutments are of different heights. 

In particular, for the un symmetrical 
arch, its closing line is not in general 
horizontal, and must be found precisely 
as that for the equilibrium polygon due 
to the applied weights. 

If, in Fig. 3, the hinge joint is not 
situated at the center, the arch is un- 
symmetrical, and the determination of 
the closing line due to the applied 
weights, is not quite so simple as in Fig. 
3. It Avill be necessary to draw the trial 
lines through the joint by which the 
curve of errors q is found. 

CHAPTER V. 

ARCH EIB AVITH END JOINTS. 

Let the curve a of the arch to be 
treated have a span of six times the rise, 
as represented in Fig. 6, and having 
divided the span into twelve equal parts, 
make the ordinates of the type bd twice 
the ordinates ab. 

Let a uniform load having a depth :cy 
cover the two-thirds of the span at the 
left, and a uniform load having a deptli 



26 



NEW CONSTEUCTIONS 



xy'^^\xy cover the one-third of the 
span at the right. Assume any pole dis- 
tance, as of one-third of the span, and 
lay off 5^2^j—aJ2/= one-half of the load 
supposed to be concentrated at the cen- 
ter; w^%o^-=^2xy^=^\hQ load concentrated 
above h^^ etc. Similarly at the left make 
5/^(j/ =03?/= one-half the load above h ; 
w^'io,^^=^2xy=:t\iQ load above ^/y '^^'^■l 
=.xy + xy' = ^xy = the load above b// 
w/to/=^xy=^th.e load above ^g', etc. 

From this force polygon draw the 
equilibrium polygon c, just as in Figs. 2 
and 3. 

Now the closing line of the equilibrium 
polygon for a straight girder with ends 
free to turn, must evidently pass so that 
the end moments vanish. Hence CgC/ 
is the closing line of the polygon c, and 
bj)/ is the closing line of the polygon d, 
drawn according to the same law. The 
remaining condition by which to determ- 
ine the bending moments is: 

2{Ma-M,)y=0 ,'. 2{May)^:E{M,y) 

which is the equation expressing the con- 
dition that the span is invariable, the 
summation being extended from end to 
end of the arch. 

This summation is effected first as in 
Figs. 2 and 3, by laying off as loads 
quantities proportional to the applied 
moments concentrated at the points of 
division of the arch, and thus finding the 
second equilibrium polygon, or deflection 
polygon of two upright girders, bent by 
these moments. 

Let us take one-fourth of each of the 
ordinates bdiov these loads, i.e. bni=^\ of 
^bd / mm^^^^b^d^, etc.: also bn, nii^, etc., 
equal to similar fractions of the ordinates 
of the curve c. Using d as the pole for 
this load, we obtain the total deflection 
bf^ on the left, and the same on the right 
(not drawn) due to the bending moments 

Similarly y^gj is the total deflection 
right and left due to the moments Me. 

]!^ow the equation of condition re- 
quires that ^^6^/= J/g. That this may 
occur, the ordinates of the polygon c 
must be elongated in the ratio of these 
deflections. To effect this, make ai=: 

iff Me ^^^ ^'j~ht\i ^11^ ^^ t^^ horizon- 
tals through i and j at a convenient dis- 
tance draw the vertical i^J^', then the 
lines ai^^ and aj^ will effect the required 
elongation, as previously explained. To 



obtain the center ordinate be, for ex- 
ample, make ai'=^bh .'. aj'=^be. To 
find the new pole o, draw bv j)arallel to 
CgCg' and vo horizontal, as before ex- 
plained. 

If ai^ cuts the load line at t^ and the 
horizontal through t^ cuts aj\ at t^, then 
the vertical through t.^ is the new position 
of the load line and tt^ is the new hori- 
zontal thrust. 

Now using o as the pole of the load 
line u^u/ etc., through t^ draw the equi- 
librium polygon starting from e. It 
must pass through b^ and bj, which tests 
the accuracy of the construction. 

The construction may now be com- 
pleted just as in Fig. 3, by doubling the 
pole distance, and finding the tangential 
thrust along the arch and the normal 
shear directly across the arch in the 
segments into which it is divided. The 
maximum thrust and tangential stress is 
obtained when the line load covers the 
entire span. 

To compute the effect of changes of 
temperature and other causes of like 
nature in producing thrust, shear, bend- 
ing moment etc., let us put the equation 
of deflections in the following form: 

„,j)^.^=2(A.^ . (D) 

'^ ^nn n \ nn n I 

This equation may perhaps put in 
more intelligible form the processes used 
in Figs. 4 and 5, and is the equation 
which should be used as the basis for the 
discussion of temperature strains in the 
arch. In equation (D) n is the number 
by which the rise of the arch must be 
divided to reduce it to bd, i.e., it is the 
scale of the vertical ordinates of the 
type bd, in Fig. 6, so that if bd was on 
the same scale as the arch itself, n would 
be unity. Again, n' is the scale of force, 
i.e., the number of tons to the inch; and 
7)1 is a number introduced for convenience 
so that any assumed pole distance ^9 may 
be used for the pole distance of the sec- 
ond equilibrium polygon. In Fig. 6, ^> 
= bd. 

We find m from the equation. 



P = 



EI 



m: 



EI 



2y7i'^n'^ 



from which m may be computed, for EI'\9> 
a certain known number of foot tons when 
the cross-section of the rib is given, 2^ is 



IN aRAPHICAL STATICS. 



27 



a number of inches assumed in the draw- 
ing, 01 and 7i' are also assumed. Now 
Dy is the number of inches by which 
the span is increased or decreased by the 
change of temperature, and tnDy is at 
once laid off on the drawing. 

The quantities in equation (D) are so 
related to each other, that the left-hand 
member is the product of the pole dis- 
tance and ordinate of the second equi- 
librium polygon, while the right-hand 
member is the bending moment pro- 
duced by the loading M-^oin', which 
loading is proportional to 31. The curve 
/"was constructed with this loading, and 
only needs to have its loads and ordi- 
nates elongated in the ratio of hf^ to 
^ mDy to determine the values of 
M-^nn' at the various points of division 
of the arch. One-half of each quantity 
is used, because we need to use but one- 
half the arch in this computation. Two 
lines drawn, as in Figs. 4 and 5, effect 
the required elongation. 

The foregoing discussion is on the im- 
plied assumption that the horizontal 
thrust caused by variation of tempera- 
ture is applied in the closing line hh^ of 
the arch, which is so evident from pre- 
vious discussions as to require no proof 
here. 

The quantity determined by the fore- 
going process is M-^nn' ■=q say, a cer- 
tain number of inches. Then M^=7in'q, 

and S=^M-^y=^n'q~—. in which —is the 
^ n n 

length of the ordinate in inches on the 

drawing at the point atwhich Jfis applied. 

The determination of the shearing and 
tangential stress induced by H is found 
by using H as the diameter of a circle, 
in which are inscribed triangles, whose 
sides are respectively parallel and per- 
pendicular to the segments of the arch, 
precisely as was done in Figs. 4 and 5. 

The whole discussion of the arch with 
end joints may be applied to an unsym- 
metrical arch with end joints. In that case, 
it would be necessary to draw a curve f 
at the right as well as f at the left, and 
the two would be unlike, as g and g' are. 
This, however, would afford no difficulty 
either in determining the stresses due to 
the loads, or to the variations of tem- 
perature. 

When the live load extends over two- 
thirds of the span, as in the Fig., the 
maximum bending moment is nearly in 



the middle of that live load, and is very 
approximately the largest which can be 
induced by a live load of this intensity, 
while the greatest moment of opposite 
sign is found near the middle of the un- 
loaded third of the span. 

If the curve of the arch were a para- 
bola instead of the segment of a circle, 
these statements would be exact and 
not approximate, as may be proved 
analytically. This matter will be fur- 
ther treated hereafter. 

CHAPTER VI. 

ARCH RIB WITH THREE JOINTS. 

Let the joints be at the center and ends 
of the arch, as seen in Fig. 7. Let the 
loading and shape of the arch be the 
same as that used in Fig. 6. Now since 
the bending moment must vanish at each 
of the joints, the true equilibrium curve 
must pass through each of the joints; 
^. e., every ordinate of the polygon c 
must be elongated in the ratio of dh to 
hh. To effect this, make di=^hh, and at 
a convenient distance on the horizontals 
through b and i draw the vertical i^ h^. 
Then the ratio lines di^ and dh^ will 

4 

enable us to elongate as required, or to 
find the new pole distance ^^, dimin- 
ished in the same ratio, by drawing the 
horizontal ti through i^. The new pole o is 
found in the same manner as in Fig. 6. 

Now with the new pole o and the new 
load line through t^ we can draw the 
polygon 6 starting at d. It must then 
pass through h^ and 5/ which tests the 
accuracy of the construction. 

The maximum thrust, and tangential 
stress is attained when the live load 
covers the entire span. 

Variations in length due to changes 
of temperature induce no bending mo- 
ments in this arch, but there may be 
slight alteration in the thrust, etc., pro- 
duced by the slight rising or falling of 
the crown due to the elongation or 
shortening of the arch. This is so small 
a displacement that it is of no import- 
ance to compute the stresses due to it. 
We have for the same reason, in the 
previous and subsequent constructions, 
omitted to compute the stresses arising 
from the displacement which the arch 
undergoes at various points by reason of 
its being bent. It would be quite pos- 
sible to give a complete investigation of 
these stresses by analogous methods. 



28 



NEW CONSTEUCTIONS 



The construction above given is appli- 
cable to any arch with three joints. The 
arch need not be symmetrical, and the 
three joints can be situated at any points 
of the arch as well as at the points 
chosen above. 

CHAPTER VII. . 

THE ARCH EIB WITH ONE END JOINT. 

Let the arch be represented by Fig. 8, 
in which the load, etc., is the same as in 
Fig. 6. 

The closing line must pass through the 
joint, for at this joint the bending 
moment vanishes. 

A second condition which must be 
fulfilled is, that the total deflection be- 
low the tangent at the fixed end of a 
straight girder having one end joint 
vanishes, for the position of the joint is 
fixed. This is expressed by the equation 

in which the summation is extended 
from end to end. 

This condition will enable us to draw 
the closing line of the polygon c, and 
also that of cl. The problem may be 
thus stated: — In what direction shall a 
closing line such as cji' be drawn from 
Cg so that the moment of the negative 
triangular area cfi^h' about Cg shall be 
equal to the moment of the positive 
parabolic area cf>c^ 

To solve this problem, first find the 
center of gravity of the parabolic area 
by taking it in parts. The parabolic 
area Cg h c/ is a segment of a single 
parabola whose area is ^hp^y^c^c^-^^h^ 
X ^6^6^ when Aj=the height of an equiva- 
lent triangle having the span for its base 

Lay off ljb=c^c^, and draw I J) J .*. 
hj,^=h^. Lay off c^p^-^h^ as proportion- 
al to the weight of the parabolic area. 
Again, c/p is proportional to the weight 
of the triangle c^c^'c^'. The parabolic 

area «= fc/c/x VV=i^^2X^eV. as 
before, .*. h^z=z^cjc^\ which may be 
found as h^ was before. 

Let h^--pp^^ then on taking any pole, 
as ^2, of this weight line, we draw qq^ \\ 
c^c^, since the left parabolic area has its 
center of gravity in the vertical through 
3'j, and the triangular area in that through 
2', we draw qq^' \\ c^p^ to the vertical 
through g/, which contains the center of 
gravity of the right parabolic area. 
The position of q midway between the 



verticals containing b and h^ is slightly 
to the right of its true position, as it 
should be at one-third of the distance 
from the vertical through h to that 
through ^2* This does not affect the 
nature of the process however. 

Then q^q^ \\ c^p^ and q^q^ \\ c^p^ give ^, 
in the vertical through the center of grav- 
ity of the total positive area. The nega- 
tive area, since it is triangular, has its cen- 
ter of gravity in the vertical through c/. 

iSTow if the total positive bending mo- 
ment be considered to be concentrated 
at its center of gravity and to act on a 
straight girder it will assume the shape 
rq^i\ of this second equilibrium polygon, 
and if a negative moment must be ap- 
plied such that the deflection vanish, the 
remainder of the girder must be r^i\, a 
prolongation of rr^. Now draw c^p^ \\ 
rr^, and we have p^p^'=^(^^h' the height 
of the triangle of negative area. Hence 
cji' is the closing line, fulfilling the re- 
quired conditions. 

Again, to draw the closing line hji^ 
according to the same law, we know 
that the center of gravity of the poly- 
gonal area d is in the center vertical. 
To find the height p^o\ of an equivalent 
triangle having a base equal to the span, 
we may obtain an approximate result, as 
in Fig. 2, by taking one twelfth of the 
sum of the ordinates of the type hd^ but 
it is. much better to obtain an exact 
result by applying Simpson's rule which 
is simplified by the vanishing of the end 
ordinates. The rule is found to reduce 
in this case to the following: — The 
required height is one eighteenth of the 
sum of the ordinates with even subscripts 
plus one ninth of the sum of the rest. 

[N'ow this positive moment concentrated 
in the center vertical and a nes^ative 
moment such as to cause no total deflec- 
tion in a straight girder, will give as a 
second equilibrium polygon ^^/^^V/; 
and if c^p^ \\ rr^', then p'p^^hfi is the 
height of the triangular negative area, 
and the closing line is hjc' , 

Now the remaining condition is that 
the span is invariable, which is expressed 
by the equation 

'2(Ma-M,)y=^, or :2[May)^^{Mcy), 

Let us construct the deflection curve 
due to the moments M^ in a manner 
similar to that employed in Fig. 2. We 
lay off quantities dm^^ ^^^5*?^4> etc., 



IN GRAPHICAL STATICS. 



29 



equal to one-fourth of the corresponding 
ordinates of the curve d, and dn^^ 
n^n^^ etc., one-fourth of the ordinates 
of the curve c. We use one-fourth or 
any other fraction or multiple of both 
which may be convenient. By using b 
for a pole we obtain the deflection curves 
^/"andy for the moments proportional to 
Md , and the curves g and g' for those 
proportional to Mc . 

Xow, Prop. IV. requires that the or- 
dinates of the polygon c should be in- 
creased so that gg' shall become equal to 
ff. Make di=^gg' and dj^ff and draw 
as before the ratio lines di^ and dj^^ then 
the vertical through t^ is the new position 
of the load line. 

Find the new length of hh which is 
he, and with the new pole o, draw the 
polygon e starting at e. It must pass 
through &g. The new pole o is found 
thus: draw bv \\ hh\ then v divides the 
weight line into two parts, which are 
the vertical resistances of the abutments. 
From %\ draw v^o \\ kk\ then the closing 
line of the polygon e has the direction Jck\ 

A single joint at any point of an un- 
svmmetrical arch can be ' treated in a 
similar manner. 

A thrust produced by temperature 
strains will be applied along the closing 
line kk\ and the bending moments in- 
duced will be proportional to the ordin- 
ates of the polygon d from this closing 
line. The variation of span must be 
computed not for the horizontal span, 
but for the projections of it on the clos- 
ing line kJi'. The construction of this 
component of the total effect will be 
like that previously employed. Another 
effect will be caused in a line perpendic- 
ular to Jck'. The variation of span for 
this construction, is the projection of the 
total horizontal variation on a line per- 
pendicular to kk'y and the bending mo- 
ments induced by this force applied at 
^g, and perpendicular to the closing line, 
will be proportional to the horizontal 
distances of the points of division from 
^g. As these constructions are readily 
made, and the shearing and tangential 
stresses determined from them, it is not 
thought necessary to give them in detail. 

CHAPTER VIII. 

ARCH RIB WITH TWO JOINTS. 

Let us take the two joints, one at the 
center and one at one end as represented 



in Fig. 9. Let the loading, etc., be as 
in Fig. 6. 

The closing line evidently passes 
through the two joints, as at them the 
bendino; moment vanishes. 

The remaining condition to be fulfilled 
is that the deflection of the right half of 
the arch in the direction of this line, 
shall be the same as that of the left 
half. 

Let us then suppose that the straight 
girder b/ p' perpendicular to the closing 
line, is fixed at b^' and bent first by 
the moments Ma giving us the deflection 
curve b/ f^ when bj is taken as the pole, 
and the loads of the type 7nm are one- 
quarter of the corresponding ordinates 
of the polygon d; and secondly, by the 
moments Mc giving us the deflection 
curve bjg' when drawn with the same 
pole, and the loads of the type 7in also 
one-quarter of the corresponding ordi- 
nates of the polygon c. It should be 
noticed that the points at which these 
moments are supposed to be concentra- 
ted in the girder b/ p' , are on the paral- 
lels to hh' through the points c?g, d^^ 
etc. 

Similarly let ^3 and/3/g be the deflec- 
tion curves of the straight girder d^p 
(using c^g as the pole distance), under the 
applied moments. 

We have used now a pole distance 
differing from that used in the right half 
of the arch. These pole distances must 
have the same ratio that the quantity EI 
has for the two parts of arch. If EI\^ the 
same in both parts of the arch the same 
pole distance must be used to obtain the 
deflection curves in both sides of the mid- 
dle. In the same manner the curves gg^ 
and ^g^g are found. Now must the mo- 
ments Mc causing the total deflection 
p' g' —gg^-=z\ai be elongated so that they 
shall cause a total deflection pf —ff^r^ 
\aj. The ratio lines ai^, aj\' will enable 
us to find the new position t^_ of the load 
line to effect this. 

To find the new pole, through 
v.„ which divides the load line into 
parts which are the vertical resistances 
of the piers, draw v^o \\ bjc. Then draw 
the polygon e as in Fig. 7, starting from 
d. It must pass through b^. We can 
find also whether he^ has the required 
ratio to hc^ by the aid of the ratio lines, 
which will further test the accuracy of 
the work. 



30 



NEW CONSTKUCTIONS 



Any unsymmetrical arch with joints 
situated differently from the case consid- 
ered can be treated by a like method. 

The temperature strains should be 
treated like those in Fig. 8, which are 
caused by a thrust along the closing line. 
Those at right angles to this line vanish 
as the joints allow motion in this direc- 
tion. The shearing and tangential stress- 
es can be found as in Fig. 3. 

Arches with more than three hinge 
joints are in unstable equilibrium, 
and can only be used in an inverted 
position as suspension bridges. These 
will be treated subsequently. If the 
joints, however, possess some stiffness 
so that they are no longer hinge joints, 
but are block-work joints, or analo- 
gous to such joints, we may still con- 
struct arches which are stable within 
certain limits although the number of 
joints is indefinitely increased. Such 
are stone or brick arches. These will 
also be treated subsequently. 

The constructions in Figs. 6, 7, 8, 9, 
can be tested by a process like that em- 
ployed in Figs. 2 and 3. In Fig. 2, for 
instance, we obtained the algebraic sum 
of the squares of the quantities of the 
type ss, and showed that such sum van- 
ishes. We can obtain the same result in 
all cases. 

CHAPTER IX. 

THE cixciNXATi a:n-d covingtox suspe:n'- 

SION BRIDGE. (Fig. 10.) 

The main span of this bridge has a 
length of 1057 feet from center to cen- 
ter of the towers, and the end spans are 
each 281 feet from the abutment to the 
center of the tower. The deflection of 
the cable is 89 feet at a mean tempera- 
ture, or about 1 — 11.87th of the span. 
There is a single cable at each side of 
the bridge. Each of these cables is made 
up of 5200 ISTo. 9 wires, each wire having 
a cross-section of 1-6 0th of a square 
inch and an estimated strength of 1620 
lbs. Each of these cables has a diameter 
of 12 J inches, and an estimated strength 
of 4212 tons. Each cable rests at the 
tower upon a saddle of easy curvature, 
the saddle being supported by 32 rollers 
which run upon a cast iron bed-plate 
8X11 feet, which forms part of the top 
of the tower. Since the bed-plate is 
horizontal this method of support ensures 
the exact perpendicularity of the force 



which the cables exert upon the towers, 
without its being necessary to make the 
inclination of the cable on both sides of 
the saddle the same. There is, there- 
fore, no tendency by the cables to over- 
turn the towers, and they need only be 
proportioned to bear the vertical stresses 
coming upon them. 

As this bridge differs greatly in some 
respects from other suspension bridges, 
it seems necessary to describe its 
peculiarities somewhat minutely. 

The roadway and sidewalks make a 
platform 36 feet wide, extending from 
abutment to abutment, 1619 feet. It is 
built of three thicknesses of plank solid- 
ly bolted together, in all 8 inches thick. 
This is strengthened by a double line of 
rolled I girders, 1630 feet long, running 
the entire length of the center of the 
platform. These I girders are arranged 
one line above the other, and across be- 
tween them, at distances of 5 feet, run 
lateral I girders which are suspended 
from the cable. The upper line of 
girders is 9 inches deep, (and 30 lbs. per 
foot); the lower line is 12 inches deep 
(and 40 lbs. per foot). The lateral 
girders are 7 inches deep (and 20 lbs. per 
foot), and are firmly embraced between 
the double line of longitudinal girders. 
The girders of this center line are 
each 30 ft. long, and are spliced together 
by plates in the hollows of the I, but 
the holes through which the bolts pass 
are slots whose length is two or three 
times the diameter of the bolts. This 
makes a " slip joint " such as is often 
used in fastening the ends of the rails on 
a railroad. The slip joints permit the 
wooden planking of the roadway to ex- 
pand and contract from variations of 
moisture and temperature without inter- 
ference from the iron girders which are 
bolted to it. 

There is also a line of wr ought-iron 
truss-work about 10 feet deep extending 
from abutment to abutment on each side 
of the roadway, consisting of panels of 
5 feet each, to each lower joint of which 
is fastened a lateral girder and a suspen- 
der from the cable. This trussing is a 
lattice, with vertical posts, and ties ex- 
tending across two panels, and its chords 
are both made with slip joints every 30 
feet. 

It is apparent that this whole arrange- 
ment of flooring with the girders and 



IX GRAPHICAL STATICS. 



31 



trusses attached to it possesses a very 
small amount of stiffness, in fact the 
stiffness is principally that of the floor- 
ing itself. It will permit a very large 
deflection, say 25 feet, up or down from 
its normal position without injury. Its 
office is something quite different from 
that of the ordinary stiffening truss of a 
suspension bridge. It certainly serves 
to distribute concentrated loads over 
short distances, but not to the extent re- 
quired, if that were the sole means of 
preserving the cable in a fixed position 
under the action of moving loads. Its 
true function is to destroy all vibrations 
and undulations, and prevent their pro- 
pagation from point to point by the 
enormous friction al resistance of these 
slip joints. When a wave does work 
against elastic forces, the reaction of 
those forces returns the wave with 
nearly its original intensity, but when it 
does work against friction it is itself 
destroved. 

The means relied on in this bridge to 
resist the effect of unbalanced loads is a 
system of stays extending from the top 
of the tower in straight lines to those 
parts of the roadway which would be 
most deflected by such loads. There are 
76 such stays, 19 from the top of each 
tower. The longest stays extend so far 
as to leave only .^50 feet., i.e., a little 
over one-third of the span, in the center 
over which they do not extend. Each 
stay being a cable 2^ inches in diameter 
has an estimated strength of 90 tons. 
They are attached every 15 feet to the 
roadway at the lower joints of the truss- 
ing, and are kept straight by being fast- 
ened to the suspenders where they cross 
them. This system is shown in Fig. 10 in 
which all the stays for one cable are 
drawn, together with every third sus- 
pender. The suspenders occur every 5 
feet throughout the bridge but none are 
shown in the figure except those attach- 
ed at the same points as the stays. 

These stays must sustain the larger 
part of any unbalanced load, at the same 
time producing a thrust in the roadway 
against either the abutment or tower. 

It is really an indeterminate ques- 
tion as to how the load is divided 
between the stays and trussing; and I 
this the more, because of the manner in i 
which the other extremities of the stays ' 
are attached. Of the nineteen stays ! 



carried to the top of one tower, the eight 
next the tower are fastened to the bed 
plate under the saddle, and so tend to 
pull the tower into the river; the remain- 
ing eleven are carried over the top of 
the tower, and rest on a small independ- 
ent saddle, beside the main saddle, and 
are eight of them fastened to the middle 
portion of the side spans as shown in Fig. 
10, while the other three are anchored to 
the abutment. 

In view of the indeterminate nature 
of the problem, it has seemed best to 
suppose that the stays should be j^ropor- 
tioned to bear the whole of any excess 
of loading of any portion of the bridge, 
over the uniformly distributed load 
(which latter is of course borne by the 
cable itself) ; and further that the truss 
really does bear some fraction of the 
unbalanced load, and that the bending 
moments have therefore the same relative 
amounts as if they sustained the entire 
unbalanced load. This fraction, how- 
ever, is quite unknown owing to the im- 
possibility of finding any approximate 
value of the moment of inertia / for the 
combined wood and iron work of the 
roadway. 

This method of treatment has for our 
present purpose this advantage, that the 
construction made use of is the same a8 
that which must be used when there are 
no stays at all, and the entire bending 
moments induced by the live loads are 
borne by the stiffness of the truss alone. 

IsTow in order to determine the tension 
in any stay, as for instance that in the 
longest stay leading to the right hand 
tower, lay off v^v^ equal to the greatest 
unbalanced weight, which under any 
circumstances is concentrated at its lower 
extremity. This weight is sustained by 
the longitudinal resistance of the floor- 
ing, and the tension of the stay. The 
stresses induced in the stay and flooring 
by the weight, are found by drawing 
from v^ and v^ the lines v^o and v.,o par- 
allel respectively to the stay and the 
flooring. Then v^o is the tension of the 
stay, and that of the other stays may be 
found in a similar manner. 

It is impossible to determine with the 
same certainty how the stress ov., paral- 
lel to the flooring is sustained. It may 
be sustained entirely by the compression 
it produces in the part of the flooring 
between the weight and the tower or the 



NEW CONSTRUCTIONS 



abutment; or it maybe sustained by the 
tension produced in the flooring at the 
left of the weight; or the stress ov^ may 
be divided in any manner between these 
two parts of the flooring, so that v^v^' 
may represent the tension at the left, 
and ov^' the compression at the right of 
the weight. It appears most probable 
that the induced stress is borne in the 
case before us by the compression of the 
flooring at the right, for the flooring is 
ill suited to bear tension both from the 
slip joints of the iron work and the want 
of other secure longitudinal fastenings; 
but on the contrary it is well designed 
to resist compression. The flooring 
must then be able at the tower to resist 
the sum of the compressions produced by 
all the unbalanced weights which can 
be at once concentrated at the extremi- 
ties of the nineteen stays. 

There is one considerable element of 
stiffness which has not been taken account 
of in this treatment of the stays, which 
serves very materially to diminish the max- 
imum stresses to which they might other- 
wise be subjected. This is the intrinsic 
stiffness of the cable itself which is formed 
of seven equal subsidiary cables formed 
into a single cable, by placing six of 
them around the seventh . central cable, 
and enclosing the whole by a substantial 
wrapping of wire, so that the entire 
cable having a diameter of 12 J inches, 
affords a resistance to bending of from 
one sixth to one half that of a hollow 
cylinder of the same diameter and equal 
cross section of metal. Which of these 
fractions to adopt depends somewhat 
on the tightness and stiffness of the 
wrapping. 

It is this intrinsic stiffness of the cable 
which is largely depended upon in the cen- 
tral part of the bridge, between the two 
longest stays, to resist the distortion 
caused by unbalanced weights. 

As might be foreseen the distortions 
are actually much greater in the central 
part of the bridge than elsewhere, though 
they would have been by far the greater 
in those parts of the bridge where the 
stays are, had the stays not been used. 

The center of a cable is comparatively 
stable while it is undergoing quite con- 
siderable oscillations, as may be readily 
seen by a simj^le experiment with a rope 
or chain. 

Let us now determine the relative 



amount of the stresses in the stiffening 
truss, on the supposition that the actual 
stresses are some unknown fraction of 
the stresses which would be induced, if 
there were no stays, and the truss was 
the only means of stiffening the cable. 
We, therefore, have to determine only 
the total stresses, supposing there are no 
stays, and then divide each stress ob- 
tained by ?i (at present unknown) to ob- 
tain the results required. Let us draw 
the equilibrium polygon d which is due 
to a uniform load of depth xy, and which 
has a deflection bd six times the central 
deflection of the cable. The loading of 
the cable is so nearly uniform, that each 
of the ordinates of the type bd, may be 
considered with sufficient accuracy to be 
six times the corresponding ordinate of 
the cable. Any multiple other than six 
might have been used with the same 
facility. In order to cause the polygon 
to have the required deflection with any 
assumed pole distance it is necessary to 
assume the scale of weights in a particu- 
lar manner, which may be determined 
easily in several ways. Let us find it 
thus : 

Let TF^one of the concentrated weights. 

Let Z>= central deflection of cable. 

Let /S'=span of the bridge 

Let J!/= central bending moment due to 
the applied weights. 

Then, if the pole distance =JaS, M=^S 
X6Z>=:2/S'Z>, for the moment is the pro- 
duct of the pole distance by the ordinate 
of the equilibrium polygon. Again, com- 
puting the central moment from the ap- 
plied forces, 

M=hh TFx iS-5 TFx i S=^ WS, 

in which the first term of the right hand 
member is moment of the resistance of 
the piers, and the second term is the mo- 
ment of the concentrated weights applied 
at their center of gravity. 

Hence, if one-third of the span is to 
represent the pole distance or true hori- 
zontal tension of an equilibrium curve 
having six times the deflection of the 
cable, each concentrated weight when 
the span is divided into twelve equal 
parts, is represented by a length equal to 
f of the deflection of the cable. The 



IN GRAHHICAL STATICS. 



33 



true horizontal tension of the cable will 
be six times that of the equilibrium 
polygon, or it will be represented, in the 
scale used, by a line twice the length of 
the span. Now taking h as the pole, at 
distances bb^=bb^=iS, lay off b/io^ = 
b^w/=iW=^D, so that they together 
represent the weight concentrated at b; 
and let w^io^= W, represent the weight 
concentrated at b^, etc. Then can the 
equilibrium polygon d be constructed by 
making dcl^ || bw^, d^d„ \\ bw^^ etc. If bd 
= 6Z> the polygon must pass through b^ 
and ^/, which tests the accuracy of the 
work. 

Xow to investigate the effect of an 
unbalanced load covering one-half the 
span, let us take one half the load on the 
right half of the span and jDlace it upon 
its left, so that x?. and xb represent the 
relative intensity of the loading upon 
the left and right half of the span re- 
spectively, the total load being the same 
as before. If it is desirable to consider 
that the total load has been increased 
by the unbalanced load we have simply 
to change the scale so that the same 
length of load line as before, (viz, b^n^^ 
+ h^vi^) shall represent the total loading. 
This will give a new value to the hori- 
zontal tension also. 

Now let a new equilibrium polygon c be 
drawn, which is due to the new distribu- 
tion of the concentrated weights. It is 
necessary to have the closing line of this 
polygon c horizontal, and this may be ac- 
complished either, by drawing the polygon 
in any position and laying off the ordi- 
nates of the type be equal to those in the 
polygon so drawn, or better as is done 
in this Figure by laying off in each 
weight line that part of the total load 
which is borne by each pier, which is 
readily computed, as follows. The 
distance of the center of gravity of the 
loading divides the span in the ratio of 
17 to 27. Hence \\ and i I of the total 
load are the resistances of the piers, or 
since the total load= 11 TFJ we have b^u^ 
= rt TFand b^^=^l W. Now make u^ 
ii^=ihe weight concentrated at b^, etc., 
and b^u^-\-b^it^=z that at b^. Then draw 
the jDolygon c. 

The polygon c has the same central 
deflection as the polygon d; for compute 
as before. 



in which the first term of the second 
member is the moment of the resistance 
of the right pier, and the second term is 
the moment of the concentrated weights 
applied at their center of gravity. 

By similar computations we may prove 
the following equalities; 



d^c=d^c^ dlc^'-----d^c^ ; 

d^c„^=—d'c\ 



d:< ; 



The quantities of the type dc are propor- 
tional to the bending moments which the 
stiffening truss must sustain if it pre- 
serves the cable in its original shape, 
when acted on by an unbalanced load 
of depth bx, on the supposition that the 
truss has hinge joints at its ends, and is 
by them fastened to the piers. For in 
that case the cable is in the condition of 
an arch with hinge joints at its ends. 
The condition which then holds is this: 



^{31ay)=^:2{Moy) 



or. 



^{Ma-Mc)y=0,\ 2{cd)y=0. 

This last is fulfilled as is seen by the 
above equations, for to every product 
such as + b^d^ X d^^c^ corresponds another 
— b/d/xd/c/ of the same magnitude 
but opposite sign. 

The polygon c could have been ob- 
tained by a second equilibrium polygon 
in a manner precisely like that used be- 
fore, but as it appears useful to show 
the connection between the methods of 
treating the arch rib which is itself stiff, 
and the fiexible arch or cable, which is 
stiffened by a separate truss, we have 
departed from our previously employed 
method for determining the polygon c, 
as it is easy to do when both c and d are 
parabolic. 

Now let us compute the bending mo- 
ment 

= d^c^^XiS=3I^-3fci 

3fc=h^WxhS=-^\^WS 

3Ici = V-i\^x 1-2 S= H WS 

.'. Mc-3fd=-h'\VS. 

Compute also the bending moment at 
the vertical through b^, 

Mc='i- Wx hS-i Wx I'^S^ WjS 

Ma=h^ Wx ^S- Wx ^S=^ WS 

.-. Mc-3ra=^WS 



34 



KEW CONSTRUCTIONS 



Similar computations may be made for 
the remaining points, and this note- 
worthy result will be found true, that 
the bending moments induced in the 
stiffening truss by the assumed loading, 
are the same as would have been induced 
by a positive loading on the left of a 
depth 2/2, and a negative loading on the 
right of an equal depth yh. For com- 
pute the moments due to such loading 
at the points h^ and 5^. 

The resistance of the pier due to such 
loading = f FT 

and 

M=. f Tfx i >Sf- J IFx hS= k WB, etc. 

We arrive then at this conception of 
the stresses to which the stiffening truss 
is subjected, viz: — the truss is loaded 
with the applied weights acting down- 
ward, and is drawn upward by a uni- 
formly distributed negative loading, 
whose total amount is equal to the posi- 
tive loading, so that the load actually 
applied at any point may be considered 
to be the algebraic sum of the two loads 
of different signs which are there applied. 
This conception might have been derived 
at once from a consideration of the fact 
that the cable can sustain only a uniform 
load, if it is to retain its shape; but it 
appears useful in several regards to show 
the numerical agreement of this state- 
ment with Prop. IV of which in fact it 
is a particular case. It is unnecessary 
to make a general proof of this agree- 
ment, but instead we will now state a 
proposition respecting stiffening trusses, 
the truth of which is sufficiently evident 
from considerations previously adduced. 

Prop. VI. The stresses induced in the 
stiffening truss of a flexible cable or arch, 
by any loading, is the same as that which 
would be induced in it by the application 
to it of a combined positive and negative 
loading distributed in the following 
manner, viz : the positive loading is the 
actual loading, and the negative loading 
is equal numerically to the positive load- 
ing, but is so distributed as to cause no 
bending moments in the cable or arch, 
i.e.^ the cable or arch is the equilibrium 
polygon for this negative loading. 



By flexible cable or arch is meant one 
which has hinge joints at the points 
where it supports the stiffening truss. It 
need not actually have hinge joints at 
these points : the condition is sufficiently 
fulfilled if it is considerably more flexi- 
ble than the truss which it supports. 

The truth of Prop. VI has been recog- 
nized by previous writers upon this sub- 
ject in the particular case of the parabolic 
suspension cable, and it has been errone- 
ously applied to the determination of the 
bending moments in the arch rib in gen- 
eral. It is inaccurate for this purpose in 
two particulars, inasmuch as in the first 
place the arch to which it is applied is 
not parabolic, though the negative load- 
ing due to it is assumed to be uniform, 
and in the second place the horizontal 
thrust is not the same for the different 
kinds of arch rib, while this assumes the 
same thrust for all, viz : that arising 
from a flexible arch or one with three or 
more joints. 

A similar proposition has been intro- 
duced into a recent publication on this 
subject*, but in that work the truss stiff- 
ens a simple parabolic cable, and the 
truss is not supposed to be fastened to 
the piers, so that it may rise from either 
pier whenever its resistance becomes 
negative. As this should not be permit- 
ted in a practical construction the case 
will not be discussed. In accordance 
with Prop. VI let us determine anew 
the bending moments due to an unbal- 
anced load on the left of an intensity 
denoted by hz. As before seen this pro- 
duces the same effect as a positive load- 
ing of an intensity yz^:^fm-=^^hz on the 
left, and a negative loading of an inten- 
sity yh^=fn=^^bz. ISTow using g as a pole 
with a pole distance of gf^ = OT\Q third of 
the span lay off the concentrated weight 
ji9^ji92=that applied at &j, etc., on the 
same scale as the weights were laid off 
in the previous construction, and in such 
a position that g is opposite the middle 
of the total load, which will cause the 
closing line to be horizontal. Then 
draw the equilibrium polygon a due to 
these weights. The ordinates of the 
type af are by Prop. VI proportional to 
the bending moments induced in the 
stiffening truss by the unbalanced load 
when the truss is simply fastened to the 

* Graphical Statics, A. J. Du Bois, p. 329, published 
by John Wiley & Son, New York. 



i:n' graphical statics. 



35 



piers at the ends, and, as we have seen, j 
each of the quantities C(/'is identical with 
the corresponding quantity cd. 

If the stiffening truss is fixed horizon- ' 
tally at its ends a closing line hli' must ' 
be drawn in such a position that ^(-^T/) 
= 0, and as it is evident that it must di- 1 
vide the equilibrium polygon symmetri- ■ 
cally it passes through / its central 
point. ! 

As stated in a previous article, the 
maximum bending moments at certain 
points of the span are caused when the \ 
unbalanced load covers somewhat more 
than half of the span. In the case of a 
parabolic cable or arch the maximum 
maximorum bending moment is caused 
when this load extends over two-thirds \ 
of the span, as is proved by Rankine in ! 
his Applied Mechanics by an analytic 
process. Let the load extend then over ■ 
all except the right hand third of the 
span with an intensity represented by ' 
hz=q^q^'. Then if fJq,=^f,'q/, the 
truss may by Prop. VI be considered to 
sustain a positive load of the intensity 
f/q^ on the left of b„\ and a negative 
load of the intensity f^^q/ on the right 
of ^/. Using g^ as the pole and the 
same pole distance as before, lay off the | 
weight q^q.^ concentrated at J., etc., so j 
that g' is opposite the middle of the ' 
weight line. We thus obtain the equili- 
brium polygon e, in which the ordinates ; 
of the type ef are proportional to the \ 
bendinsc moments of the truss under the 
assumed loading, when its ends are sim- 
ply fastened to the piers. 

Now bd was the ordinate of an equili- 
brium polygon having the same horizon- 
tal tension, and under a load of the same 
intensity covering the entire span. It 
will be found that bd=^/\e^, which may 
be stated thus: — the greatest bending 
moment induced in the stiffening truss, 
by an unbalanced load of uniform in- 
tensity is four twenty-sevenths of that 
produced in a simple truss under a load 
of the same intensitv coverins^ the entire 
span. This result was obtained by Ran- 
kine analytically. If the truss is fixed 
horizontally at its ends, we must draw a 
closing line kk', which fulfills the condi- 
tions before i^sed for the straight girder 
fixed at the ends, as discussed previously 
in connection with the St. Louis Arch. 
By the construction of a second equili- 
brium polygon, as there given, we find 



the position of Jck'; then the ordinates 
ke will be proportional to the bending 
moments of the stiffenins; truss. 
The shearing stress in the truss is obtained 
from the loading which causes the bend- 
ing moment, in the same manner as that 
in any simple truss. The horizontal ten- 
sion in the cable, is the same whenever 
the total load on the span is the same, 
and is not changed by any alteration in 
the distribution of the loading, which 
fact is evident from Prop. VI. The 
maximum tension of the cable is found 
when the live load extends over the 
entire span, and is to be obtained from a 
force polygon which gives for its equili- 
brium polygon the curve of the cable 
itself, as would be done by using the 
weights w^w^, etc., and a pole distance of 
six times ^;^^=twice the span. 

The temperature strains of a stiffening 
truss of a suspension bridge are more 
severe than those of the truss stiffening 
an arch, because the total elongation of 
the cable in the side spans as well in the 
main span, is transmitted to the main 
span and produces a deflection at its 
center. This is one reason why stays 
furnish a method of bracing, particularly 
applicable to suspension bridges. But 
supposing that the truss bears part of 
the bending moment due to the elonga- 
tion of the cable, it is evident that when 
the truss is simply fastened to the piers, 
the bending moments so induced are 
proportional to the ordinates of the type 
bd, for by the elongation of the cable, it 
transfers part of its uniformly distrib- 
uted weight to the truss. 

That load which the cable still sus- 
tains, is uniformly distributed, if the 
cable still remains parabolic, therefore 
that transferred to the truss is uniformly 
distributed. 

When the truss is fixed horizontally 
at the piers, the closing line of the curve 
d must be changed so that 2{M)=^0, 
and the bending moments induced by 
variations of temperature, will be pro- 
portional to the ordinates between the 
curve d and this new closing line. 

It remains only to discuss the stability 
of the towers and anchorasje abutments. 
The horizontal force tending to overturn 
the piers comes from a few stays only, 
as was previously stated, and is of such 
small amount that it need not be consid- 
ered. 



36 



:kew constructions 



The weight of the abutment in 
the case before us is almost exactly 
the same as the ultimate strength 
of the cable. Suppose that st=sv are 
the lines representing these quantities in 
their position relatively to the abutment. 
Since their resultant sv intersects the 
base beyond the face of the abutment, 
the abutment would tip over before the 
cable could be torn asunder. And since 
the angle vsr is greater than the angle 
of friction between the abutment and 
the ground it stands on, the abutment if 
standing on the surface of the ground, 
would slide before the cable could be 
torn asunder. 

The smallest value which the factor of 
safety for the cable assumes under a 
maximum loading is computed to be six. 
Take sf=ist as the greatest tension 
ever induced in the cable, then sr' the 
resultant of sv and sf cuts the base so 
far within the face that it is apparent 
that the abutment has sufficient stability 
against overturning, and the angle VS7'' 
is so much smaller than the least value 
of the angle of friction between the 
abutment and the earth under it, that 
the abutment would not be near the 
point of sliding even if it stood on the 
surface of the ground. It should be 
noticed tha all the suspenders in the 
side span assist in reducing the tension of 
the cable as we approach the abutment, 
and conduce by so much to its stability. 
Also the thrust of the roadway may as- 
sist the stability of the abutment, both 
with respect to overturning and sliding. 

CHAPTER X. 

THE CONTINUOUS GIRDER WITH VARIABLE 

CROSS-SECTION. 

In the foregoing chapters the discussion 
of arches of various kinds has been shown 
to be dependent upon that of the straight 
girder; but as no graphical discussion has, 
up to the present time, been published 
which treats the girder having a variable 
cross-section and moment of inertia, our 
discussion has been limited to the case of 
arches with a constant moment of iner- 
tia. 

Certain remarks were made, however, 
in the first chapter tending to show 
the close approximation of the i*esults 
in case of a constant moment of inertia 
to those obtained when the moment of 
inertia is variable. We, in this chapter, 



propose a new solution of the continuous 
girder in the most general case of varia- 
ble moment of inertia, the girder resting 
on piers having any different heights 
consistent with the limits of elasticity of 
the girder. This solution Avill verify the 
remarks made, and enable us easily to see 
the manner in which the variation of the 
moment of inertia affects the distribution 
of the bending moments, and by means 
of it the arch rib with variable moment 
of inertia can be treated directly. 

Besides the importance of the con- 
tinuous girder in case it constitutes the 
entire bridge by itself, we may remark 
that the continuous girder is peculiarly 
suited to serve as the stiffening truss of 
any arched bridge of several spans in 
which the arches are flexible. Indeed, it 
is the conviction of the writer that the 
stiff arch rib adopted in the construction 
of the St. Louis Bridge was a costly mis- 
take, and that, if a metal arch was desir- 
able, a flexible arch rib with stiffening 
truss was far cheaper and in every way 
preferable. 

Let us write the equation of deflections 
in the form 

mn n \nn n I 

in which n is the number by which any 
horizontal dimension of the girder must 
be divided to obtain the corresponding 
dimension in the drawing, n' is the 
divisor by which force must be divided 
to obtain the length by which it is to 
be represented in the drawing, m is an 
arbitrary divisor which enables us to 
use such a pole distance for the second 
equilibrium polygon as may be most 
convenient, I^ is the moment of inertia 
of the girder at any particular cross sec- 
tion assumed as a standard with which 
the values of / at other cross sections 
are compared, and z^Z^-r-/ is the ratio 
of I^ (the standard moment of inertia), 
to I (that at any other cross-section). 
For the purpose of demonstrating the 
general properties of girders, the equation 
need not be encumbered with the coeffici- 
ents mnn\ but for purposes of explaining 
the graphical construction they are very 
useful, and can be at once introduced in- 
to the equation when needed. 
In the equation 



IIS" GRAPHICAL STATICS. 



37 



the quantity D is the deflection of any 
point of the girder below the tangent 
at the point a where the summation be- 
gins, and M is the actual bending mo- 
ment at any point between and a. 
These moments M at any point consist 
in general of three quantities, represented 
in the construction by the positive ordi- 
nate of the equilibrium polygon due to 
the weights, and by the two negative ordi- 
nates of the triangles into which we have 
divided the negative moment area. If 
we distinguish these components of JSI 
by letting M^ represent that due to the 
weights, while J/, and ilf, represent the 
components due to the left and right 
negative areas respectively, the equation 
of deflections becomes 

D . EI,^ll{MJ.x)-:l{M^ix)-Il(MJx) 

Now let us take at a pier at one end 
of a span and extend the summation 
over the entire span. 




If the piers are b and h' as in Fig. 11, 

let us suppose that coincides with h 
and a with h' \ also suppose for the in- 
stant that -/is constant, so that i=\ at 
all points of the girder. Then we have 



in which />& is the deflection of b below 
the tangent at b', x^ is the distance of 
the center of gravity of the moment 
area due to the applied weights from b, 
while x^ and x^ are the distances of the 
centers of gravity of the negative areas 
from b. In Fig. 11 let cc/ be the posi- 
tive area due to the weights and repre- 



senting ^l {MX while ^j,(J/,) and J^^, 

(il/J are represented by hcc^ and AAV/ 
respectively. Let the center of gravity 
of cc^c' be in gq^^, while the centers of 
the two negative areas are in tr and t'r'. 
Let the height of a triangle on some as- 
sumed base, and equivalent in area to 
cCqC', be 7i\, then by a process like that 
in Fig. 2 it is evident that 7'7\ and r r 
are the heights of the right and left 
negative triangles, having the assumed 
base, on the supposition that the girder 
is fixed horizontally over the piers. 

Now introducing the constants 7n7i?i' 
into the last equation and into the equa- 
tion before that, the relation of the quan- 
tities is such that if the moments be ap- 
plied as weights at their centers of 
gravity with the pole distance ji9^=^/-^ 
7nn^7i\ the equilibrium polygon so obtain- 
ed will be tangent at the piers to the ex- 
aggerated deflection curve obtained when 
the distributed moments are used as 
weights; and the deflection at the pier 
b from the tangent at b' will be the same 
as that of this exaggerated deflection 
curve, and vice versa. 

Let pm—i\T,^^ p'm'—rr^ and pt:=^p't^ 
then t and t' constitute the pole, pm and 
p'm' the negative loads, and ptn-^p'm' 
the positive load. Then is btqt'h' the 
equilibrium polygon for these loads. 
The deflection of b below b't' vanishes 
as it should in case the girder is fixed 
horizontally over the pier. 

Now let the direction of the tangents 
at the piers be changed so that the 
tangents to the exaggerated deflection 
curve assume the directions bt^ and h't^. 
Then the load line and force polygon 
assume a new position, such that t^ and t ' 
form the pole, and chi^^pm and d'7i'=z 
p'm' comprise the positive load while 
np^ and n'p^ are the new negative loads 
which will cause the equilibrium polygon 
bt^qj}^'b\ which is due to them, to have 
its sides bt^ and b't^' in the directions as- 
sumed. 

There are several relations of quanti- 
ties in this figure to which we wish to 
direct attention. It is evident, in 
case / is not constant, that from the 
area cc^c' whose ordinates are propor- 
tional to _/I/g, the actual bending mo- 
ments due to the weights, another area 
whose ordinates are proportional to 



38 



NEW co:n'structions 



MJ,^ the effective bending moments, can 
be obtained by simple multiplication, 
since lis known at. every point of the 
girder. Moreover, the vertical through 
the center of gravity of this positive 
effective moment area can be as readily 
found as that through the actual positive 
moment area. Call this vertical "the 
positive center vertical." Again, the 
negative moment areas proportional to 
MJ, and MJ, can be found from the tri- 
angular areas proportional to J/^ and M, 
by simple multiplication, and if Tve pro- 
ceed to find the verticals through their 
centers of gravity we shall obtain the 
same verticals whatever be the magni- 
tude of the negative triangular areas, 
since their vertical ordinates are all 
chano;ed in the same ratio bv assumino^ 
the negative areas differently. Let us 
call these verticals the "left" and 
*' right " verticals of the span. In case 
i=l, as in Fig. 11, the left and right 
verticals divide the span at the one-third 
points. This matter will be treated 
more fully in connection with Fig. 13. 

Again, let us call the line tj}^' " the 
third closing line." It is seen that, 
whatever may be the various positions 
of the tangent ht^^ the ordinate dn^ be- 
tween the third closing line and t^q^ pro- 
longed, is invariable; for the triangle 
t^q^t^ is invariable, being dependent on 
the positive load and pole distance alone. 
By similarity of triangles it then follows 
that the ordinate, such as lo\ on any as- 
sumed vertical continues invariable; and 
when there is no negative load at ^^, 
then ht^q^ becomes straight, o' coincides 
with h and n with ^:>^. Similar relations 
hold at the right of q^. The quantity 
d}^^ is of the nature of a correction to be 
subtracted from the negative moment 
when the s^irder is fixed horizontally at 
the piers in order to find the negative 
moment when the tangent assumes a new 
position, for np^-=dn—dp^. The negative 
moments can consequently be found from 
the third closing line and the tangents 
at the piers; while the remaining lines 
qf^ and q^t^ will test the correctness of 
the work. Before applying these pro- 
perties of the deflection polygon and its 
third closing line to a continuous girder, 
it is necessary to prove a geometrical 
theorem from Fig. 12. 

Let the variable triangle xyz be such 
that the side x?. always passes through . 




the fixed point g^ the side xy always 
passes through the fixed point jl>, and the 
vertices xyz are always in the verticals 
through those points; then by the prop- 
erties of homologous triangles the side 
yz also has a fixed point/" in the straight 
line gp. Furthermore, if there is a point 
z' in the vertical through s, and in all 
positions of z it is at the same constant 
distance from z^ then on the line yz' there 
is a fixed point g' where the vertical 
through f intersects yz' ; for, if z' main- 
tains its distance zz' invariable, then 
must any other point as g' remain con- 
stantly at the same vertical distance 
from/", as appears from similarity of tri- 
angles. But as f is fixed g' is also. 
When, for instance, the triangle xyz as- 
sumes the position x^y^z^^ then z' moves 
to 2/. 

Let us now apply the foregoing to the 
discussion of a continuous girder over 
three piers 79"^;^:)' as shown in Fig. 13, 
in which the lengths of the spans have 
the ratio to each other of 2 to 3. Divide 
the total length of the girder into such a 
number of equal parts or panels, say 15, 
that one division shall fall at the inter- 
mediate pier, and let the number of lines 
in any panel of the type aa represent its 
relative moment of inertia. Assume the 
moment of inertia where there are three 
lines, as at a, «^, etc., as the standard or 
jTq, then 2=1 at a, i=^ at a^, i=^ at «/, 
etc. 

Let the polygons c and c' be those due 
to the weights in the left and right spans 
respectively. Then the ordinates of 
the type be are proportional to JI^ in the 
left span. The figure hc^c" c^'c"c^c^c^c" 
5g is the positive effective moment area 
in the left span, and its ordinates are 
proportional to MJ^. Its center of gravi- 
ty has been found, by an equilibrium 



IN GRAPHICAL STATICS. 



39 




polygon not drawn, to lie in the positive 
center vertical qq^^. A similar positive 
effective moment area on the right has 
its center of [gravity in the positive cen- 
ter vertical q q,'. 

Now assume any negative area, as 
that included between the lines h and c?, 
and draw the lines fih^ and A^/, dividing 
the negative area in each span into right 
and left triangular areas. Let the quan- 
tities of the type hh be proportional to 
if,, Ac? to M^, h'h' to J^^/, etc., then the 
ordinates of hbJj^'b^b^'hJ)Jj^'hJ, are pro- 
portional to MJ., and the center of gravi- 
ty of this area has been found to lie in 
the i-ight negative vertical t^'i\. Similar- 
ly, the left negative vertical containing 
the center of gravity of the left negative 

the right 
are the left and right 
Tcrticals. As before stated, these verti- 
cals would not be changed in position 
by changing the position in any manner 
whatever of the line d by which the 
negative moments were assumed, for 
sucPi change of position would change 
all the ordinates in the same ratio. 

Let us find also the vertical containing 
the center of gravity of the effective 
moment area, corresponding to the actual 
moment area hjih,'. It is found by a 
polygon not drawn to be vo. Call vo 
"the negative center vertical." It is 
unchanged by moving the line d. If a 



effective moments, is t.^i\. In 
span ^/r/ and t. 



V/ 



polygon be drawn due to the effective 
moments as loads, two of its sides must 
intersect on -yo, because it contains the 
center of gravity of contiguous loads. 
Now let rr^ represent 2(M,i) : — it is in 
fact one eighth of the sum of the ordi- 
nates 6jC, 4-'^jC/', etc., and hence is the 
height of a triangle having a hane^^bb^, 
and an area equal to the effective mo- 
ment area in the left span. Also r'r/ is 
the height of a triangle having the same 
base, and an area equal to the effective 
moment area in the right span. 

As previously explained, sr^ is the 
amount of the right negative effective 
moment area in the left span, measured 
in the same manner, while sr is that on 
the left when the girder is fixed horizon- 
tally at the piers. We obtain s'r/ and 
s'r' in the right span, in a similar manner. 
Xow assume the arbitrary divisor rn=l, 
and take the pole distance r^9i^=JEJI^-^ 
nW. Then as seen previously, if mn^=sr^j 
ou is the constant intercept on the nega- 
tive center vertical, between the third 
closing line in the left span, and a side 
of the type qt. Also ou' is a similar 
constant intercept on this vertical due 
to the right span. Make r^n^=r/t^ and 
n,//22=:.sr, then Ib^ is a similar invariable 
intercept; as is I'bJ, which is obtained 
in a similar manner. 

Now the negative center vertical ov 
was obtained from the triangle bJJj,^^ i.e. 



40 



NEW CONSTRUCTIONS 



on the supposition that the actual mo- 
ment over the pier is the same whether 
it be determined from the left or right 
of the pier. It is evident that while the 
girder is fixed horizontally at the inter- 
mediate pier, the moment at that pier is 
generally different on the two sides, at 
points iufinitesimally near to it, but that 
when the constraint is removed an equali- 
zation takes place. 

Since oic and oit' are derived from 
the positive effective moments, it appears 
that when the tangent at ^9 is in such a 
position that the two third closing lines 
intercept a distance uu' on ov and the 
two lines of the type qt when prolonged 
intersect on ov, the moments over the 
pier will have become equalized. 

We propose to determine the position 
of the tangent at 2^ which will cause this 
to be true, by finding the proper position 
of the third closing lines in the two spans. 

Move the invariable intercepts to a 
more convenient position, by making 
o^s=ow, and o^z'=^ou'. Now by making 
the arbitrary divisor m=l, as we did, 
the ordinates of the deflection polygon 
became simply Z>, i.e., they are of the 
same size in the drawing as in the girder, 
hence the difference of level of p", p and 
p)' niust be made of the actual size. By 
changing m this can be increased or 
diminished at will. 

Now we propose to determine two 
fixed points g and g\ through which the 
third closing line in the left span must 
pass, and similarly g'" and g' on the 
right. 

If the girder is free at p" then as shown 
in connection with Fig. 1], the third 
closing line must pass through g, if gp"=^ 
Ib^. Draw gz as a tentative position of 
the third closing line, and complete the 
triangle xy'z as in Fig. 12. 

Then is xy' the tentative position of 
the tangent at p, and since the third clos- 
ing line in the right span must pass 
through y\ and make an intercept on 
the negative center vertical equal to uu', 
then z'y' is its corresponding tentative 
position. But wherever gz may be 
drawn, every line making an intercept 
=iuit' and intersecting t^'r^' in such a 
manner that the tangent passes through 
p) must pass through the fixed point g', 
found as described in Fig. 12. There- 
fore the third closing line in the right 
span passes through g'. Similarly, if 



there were more spans still at the right 
of these, we should use g' for the deter- 
mination of another fixed point, as we 
have used g to determine it. 

Now find g'" and g" precisely as g and 
g' have been found, and draw the third 
closing lines tj;^ and t^'t^'. If tj^^' passes 
through p the construction is accurate. 
Make uu"=^vv", then is n^m^ the nega- 
tive effective moment at the left, and 
n^m^ that at the right of the pier. 

Let hw be the effective moment area 
corresponding to the triangle hM^, and 
measured in the same manner as the 
positive area was, by taking one eighth 
of its ordinates, and let hw^^=^n^m^\ then 
as the effective moment hw is to the 
actual moment hh corresponding to it, so 
is the effective moment hw^ or n^m^ to 
the actual moment hh corresponding to 
it. The same moment hh is also found 
from n^'m^', by an analogous construc- 
tion at the right of h, which tests the ac- 
curacy of the work. 

Several other tests remain which we 
will briefly mention. 

Prolong p)"t^ to q, and ^9^^/ to q', then 
qt^ and q't^' must intersect on the nega- 
tive center vertical at o^ so that o^v^'^:^ 
ou" . Also vv' must be equal to uu' . 
Again t^-o' passes through f, and ^/v 
through f. Also yo^ intersects qo^ on 
the fixed vertical /^'^ at e, and y'o^ inter- 
sects q^o^ on the fixed vertical f'g' at e' . 
That these must be so is evident from a 
consideration of what occurs during a 
supposed revolution of the tangent t^t-l, 
to the position xy' . 

Now having determined the moment 
hh over the pier, hh^ and hh^ are the 
true closing lines of the moment poly- 
gons G and g' . Call these closing 
lines h, then the ordinates of the 
type he will represent the bending mo- 
ments at different points of the girder. 
The points of the contra flexure are at 
the points where the closing lines inter- 
sect the polygons g and g' . The direc- 
tions of the closing lines will permit at 
once the determination of the resistances 
at the piers and the shearing stresses at 
any point. 

The particular difference between the 
construction in case of constant and of 
variable moment of inertia, is seen to be 
in the positions of the center verticals 
positive and negative, and the right and 
left verticals. 



IN GRAPHICAL STATICS. 



41 



The small change in their position due 
to the variation in the moment of inertia, 
is the justification of the remarks previ- 
ously made respecting the close approxi- 
mation of the two cases. 

It is seen that the process here devel- 
oped can be applied with equal facility 
to a girder with any number of spans. 
Also if the moment of inertia varies con- 
tinuously instead of suddenly, as assumed 
in Fig. 13, the panels can be taken short 
enough to approximate with any re- 
quired degree of accuracy to this case. 

CHAPTER XI. 

THE THEOREM OF THREE MOMENTS. 

The preceding construction has been 
in reality founded on the theorem of 
three moments, but when the equation 
expressing that theorem is written in 
the usual manner, the relationship is 
difficult to see. Indeed the equation as 
given by Weyrauch* for the girder hav- 
ing a variable moment of inertia, is of so 
complicated a nature that it may be 
thought hopeless to attempt to associate 
mechanical ideas with the terms of the 
equation, in any clearly defined relation- 
ship. We propose to derive and express 
the equation in a novel manner, which 
will at once be easy to understand, and 
not difficult of interpretation in connec- 
■^tion with the preceding construction. 

Let us assume the general equation of 
deflections in the form. 

I):=:E(3Ix^EI), or D.ET^:2{Mix) 

in which Z is the variable moment of 
inertia, I^ some particular value of J" as- 
sumed as the standard of comparison, 
i^I^-^I^ and x is measured horizontally 
from the point as origin, where the de- 
flection D is taken to the point of appli- 
cation of the actual bending moment 31. 
The quantity 3Ii is called the effective 
bending moment, and the deflection JD 
is the length of the perpendicular from 
the origin to the line tangent to the de- 
flection curve at point to which the sum- 
mation is extended. 

Xow consider two contiguous spans 
of a continuous girder of several spans, 
and let aci denote the piers, c being the 
intermediate pier. Let the span ac=-l 
and hc^^V. Take the origin at a and 

* Allgemeine Theorie und Berechnuug der Coutinuir- 
lichen und Einfachen Trager. Jakob I. Weyrauch. 
Leipzig 18T3. 



extend the summation to c, calling the 
deflection at a, Da- When the origin is 
at b and the summation extends to c, let 
the deflection be 2>i). Let also 2/a,2/b and 
2/c be the heights of a, h and c respective- 
ly above some datum level. Then, as 
may be readily seen, 

^a = Va — Vc — Itc , 

^b — yb — yc — Vtc, 

if tc is the tangent of the acute angle at 
c on the side towards a between the tan- 
gent line of the deflection curve at c 
and the horizontal, and ^c' is the tangent 
of the corresponding acute angle on the 
side of c towards h. 

Now if we consider equation (7) to 
refer to the span /, the moment M may 
be taken to be made up of three parts, 
viz: — M^ caused by the weights on the 
girder, M^ dependent on the moment 
Mc at c, and M^ dependent on the mo- 
ment Ma at a. The moments in the 
span V may be resolved in a similar man- 
ner. We may then write the equations 
of deflections in the two spans when the 
summation extends over each entire span 
as follows: 

ElXya-yc-U^=^l {M^x)-X (Mjx) 

-X{MJx) (8) 

^I.{yb -y-Vt:)=X{M:i'x') 

-^\{M^i'x')-^\{JSi:i'x') (9) 

in which x is measured from a, and x' 
from h towards a. Now if the girder is 
originally straight, tc =^ ~ tc ^ hence 
we can combine these two equations so 
as to eliminate tc and tc\ and the result- 
ing equation will express a relationship 
between the heights of the piers, the 
bending moments (positive and negative), 
their points of application and the mo- 
ments of inertia; of which quantities the 
negative bending moments are alone un- 
known. The equation we should thus 
obtain would be the general equation 
of which the ordinary expression of the 
theorem of three moments is a particular 
case. Before we write this general 
equation it is desirable to introduce cer- 
tain modifications of form which do not 
diminish its generality. Suppose that 

then is x^ the distance from a to the cen- 
ter of gravity of the negative effective 



42 



NEW CONSTRUCTIONS 



moment area next to c. As was shown 
in connection with Fig. 13, the position 
of this center of gravity is independent 
of the magnitude of M^ or Mq and may 
be found from the equation, 



i^nJU \XinJU 



93, 



/a . • • • 



(10) 



for M^ is proportional to x. Similarly 
it may be shown that 



X- 



i{l—x)xdx 

/a 
i{l—x)dx 



(11) 



is the distance of the center of gravity 
of the negative effective moment area 
next to a. 

Again, suppose that 

then is ^\ an ' average value of ^ for the 
negative effective moment area next to 
0, which is likewise independent of the 
magnitude of il/^, as appears from reason- 
ing like that just adduced respecting x^. 
Hence ^\ may be found from the equation 



/: 



iiXCuX 



^,= 



/a 
xdx 

Similarly it may be shown that 

/a 
i(l—x)dx 



(12) 



^„ = 



/a 
{l—x)dx 



(13) 



in which i^ is the average value of i for 
the negative effective moment area next 
to a. 

The integrals in equations (lO), (11), 
(12). (13), and in others like them refer- 
ring to the span I', which contain i must 
be integrated differently, in case i is dis- 
continuous, as it usually is in a truss, 
from the case where i varies continuous- 
ly. When i is discontinuous the integral 
extending from c to a must be separated 
into the sum of several integrals, each of 
which must extend over that portion of 
the span I in which i varies continuously. 

Furthermore we have 



X{M;}=iMj , 



(14) 



since each member of this equation rep- 



resents the negative actual moment area 
next to e in the span I. 

Similarly, we have the equations 

If there is no constraint at the pier 
then must Mo = M^' . 

Now making the substitutions in equa- 
tions (8) and (9), which have been indi- 
cated in the developments just com- 
pleted, and then eliminating tc and tc\ 






V 

^Mw^i^'\ . . . (15) 

in which "^^ is the distance from a of the 
center of gravity of the positive effect- 
ive moment area due to the weights in 
the span ^, and "^Z is a similar distance 
from h in the span l\ while i^ and ^/ are 
average values of i for these areas de- 
rived from the equations in each span, 

^•=2(i!/,^)-2(ilf,). 

It may frequently be best to leave the 
expressions containing the positive mo- 
ments in their original form as expressed 
in equations (8) and (9). 

Equation (15) expresses the theorem of 
three moments in its most general form. 

Let us now derive from equation (15), 
the ordinary equation expressing the 
theorem of three moments, for a girder 
having a constant cross section. In this 
case ^=l, and we wish to find the value 
of the term ^{M^x) in each span. Let 
M^ be caused by several weights P ap- 
plied at distances z from a, then the mo- 
ment due to a single weight P at its 
point of application is 

M^ = Pz{l-z)~l, 

which may be taken as the height of the 
triangular moment area whose base is I 
which is caused by P. This triangle 
whose area is ^Mzl is the component of 
2(M^ due to P and can be applied as a 
concentrated bending moment at its cen- 
ter of gravity at a distance x from a. 

Now x^=-^{l+z)y and taking all the 
weights P at once 

Also in equation (15) we have in this 
case 



IHq- GRAPHICAL STATICS. 



43 



QEI 



.17 - 27 —' 17' ~' 27' 



Va — Vc . 2/j>-yc 



= Mal+2Mc{l+l')-\-Mj,V . (16) 

Equation (16) then expresses the the- 
orem of three moments for a girder hav- 
ing a constant moment of inertia /, and 
deflected by weights applied in the span 
I at distances z from a, and also by 
weights in the span I' at distances z' from 
b. 

Let us also take the particular case of 
equation (15) when the moment of inertia 
is invariable and the piers on a level; then 
i=\, and if we let A^ and A J be the 
positive moment areas due to the weights 
we have 

Mal+2Mc{l-\-l')+M^V . . {11) 

This form of the equation of three mo- 
ments was first given by Greene.* 

The advantage to be derived in discus- 
sing this theorem in terms of the bending 
moments, instead of the applied weights 
is evident both in the analytical and the 
graphical treatment. The extreme com- 
plexity of the ordinary formulae arises 
from their being obtained in terms of 
the weights. 

In order to complete the analytic solu- 
tion of the continuous girder in the gen- 
eral case of equation (15), it is only 
necessary to use the well known equa- 
tions, 

M=Me+SoZ-Sl{Pz:) . . 08) 

Se=]iMa-Me+K{Pz)-] . . (l9) 

Sc' = \{M^-Mc+ll{Pz')^ . (20) 

P^ = Sc-\- jS/ (21) 

S=Se-K{P) (22) 

In (18) 3f is the bending moment at 
any point in the span I, Sc is the shear 
at c due to the weights in the span I, 
and z^ is the distance from towards c 
of the applied forces P and jSc in the seg- 
ment Oc. 

* Graphical Method for the Analysis of Bridge Trusses. 
Chas. E, Greene. Published by D. Van Nostrand. New 
York, 1875. 



Equation (19) is derived from (18) by 
taking at a, and (20) is obtained simi- 
larly in the span I'. Pc is the reaction 
of the pier at c. JS is the shear at in 
the span I. These equations also com- 
plete the solution of the cases treated in 
(16) and (11), 

CHAPTER XII. 

THE FLEXIBLE ARCH EIB AND STIFFENING 
TRUSS. 

Whenever the moment of inertia of 
an arch rib is so small, that it cannot 
afford a sufficient resistance to hold in 
equilibrium the bending moments due 
to the weights, it may be termed a flexi- 
ble rib. 

It must have a sufficient cross section 
to resist the compression directly along 
the rib, but needs to be stiffened by a 
truss, which will most conveniently be 
made straight and horizontal. The rib 
may have a large number of hinge joints 
which must be rigidly connected with 
the truss, usually by vertical parts. It 
is then perfectly flexible. 

If, however, the rib be continuous 
without joints, or have blockwork joints, 
it may nevertheless be treated as if per- 
fectly flexible, as this supposition will 
be approximately correct and on the side 
of safety, for the bending moments in- 
duced in the truss will be very nearly as 
great as if the rib were perfectly flexible, 
in case the same weight would cause a 
much greater deflection in the rib than 
in the truss. It will be sufficient to 
describe the construction for the flexible 
rib without a figure, as the construction 
can afford no difficulties after the con- 
structions already given have been mas- 
tered. 

Lay off on some assumed scale the 
applied weights as a load line, and let 
us call this vertical load line ww\ 
Divide the span into some convenient 
number of equal parts by verticals, 
which will divide the curve a of the rib 
into segments. From some point b as a 
pole draw a pencil of rays parallel to the 
segments of a, and across this pencil 
draw a vertical line im% at such a dis- 
tance from b that the distance taf/ be- 
tween the extreme rays of the pencil is 
equal to ww\ Then the segments of 
im' made by the rays of the pencil are 
the loads which the arch rib would sus- 



44 



NEW CONSTRUCTIONS 



tain in virtue of its being an equilibrium 
polygon, and they would induce no bend- 
ing moments if applied to the arch. 
The actual loads in general are different- 
ly distributed. By Prop. VI the bending 
moments induced in the truss are those 
due to the difference between the weight 
actually resting on the arch at each 
point, and the weight of the same total 
amount distributed as shown by the 
segments of the line %iu' . 

Now lay off a load line vv' made up 
of weights which are these differences 
of the segments of uu' and %ov:)\ taking 
care to observe the signs of these dif- 
ferences. The algebraic sum of all the 
weights 'C'o' vanishes when the weights 
which rest on the piers are included, as 
appears from inspection of the construc- 
tion in the lower part of Fig. 10. The 
construction above described will differ 
from that in Fig. 10 in one particular. 
The rib will not in general be parabolic, 
and the loads which it will sustain in 
virtue of its being an equilibrium poly- 
gon will not be uniformly distributed, 
hence the differences which are found as 
the loading of the stiffening truss do 
not generally constitute a uniformly 
distributed load. 

The horizontal thrust of the arch is 
the distance of %iu' from h measured on 
the scale on which the loads are laid off, 
and the thrust along the arch at any 
point is length of the corresponding ray 
of the pencil between h and ilu' . These 
thrusts depend only on the total weight 
sustained, while the bending moments 
of the stiffening truss depend on the 
manner in which it is distributed, and 
on the shape of the arch. 

Having determined thus the weights 
applied to the stiffening truss, it is to be 
treated as a straight girder, by methods 
previously explained according to the 
way in which it is supported at the 
piers. 

The effect of variations of temperature 
is to make the crown of the arch rise 
and fall by an amount which can be 
readily determined with sufficient exact- 
ness, (see Rankine's Applied Mechanics 
Art. 169). This rise or fall of the arch 
produces bending moments in the stiffen- 
ing truss, which is fastened to the tops 
of the piers, which are the same as would 
be produced by a positive or negative 
loading, causing the same deflection at 



the center and distributed in the same 
manner as the segments of uu' : for it' 
is such a distribution of loads or pres- 
sures which the rib can sustain or pro- 
duce. A similar set of moments can be 
induced in the stiffening truss by length- 
ening the posts between the rib and 
truss. 

When this deflection and the value of 
El in the truss are known, these mo- 
ments can be at once constructed by 
methods like those already- employed. 
A judicious amount of cambering of this 
kind is of great use in giving the struc- 
ture what may be called "initial stiff- 
ness." The St. Louis Arch is wanting in 
initial stiffness to such an extent that 
the weight of a single person is sufficient 
to cause a considerable tremor over an 
entire span. This would not have been 
possible had the bridge consisted of an 
arch stiffened by a truss which was an- 
chored to the piers in such a state of 
bending tension as to exert considerable 
pressure upon the arch. This tension of 
the truss would be relieved to some ex- 
tent during the passage of a live load. 

The arch rib with stiffening truss, is a 
form of which many wooden bridges 
were erected in Pennsylvania in the 
earlier days of American railroad build- 
ing, but its theory does not seem to have 
been well understood by all who erected 
them, as the stiffening truss was itself 
usually made strong enough to bear the 
applied weights, and the arch was added 
for additional security and stiffness, 
while instead of anchoring the truss to 
the piers and causing it to exert a pres- 
sure on the arch, a far different distribu- 
tion of pressures was adopted. Quite a 
number of bridges of this pattern are 
figured by Haupt* from the designs of 
the builders, but most of them show by 
the manner of bracing near the piers 
that the engineers who designed them 
did not know how to take advantage of 
the peculiarities of this combination. 
This further appears from the fact, that 
the trussing is not usually continuous. 

A good example, however, of this 
combination constructed on correct prin- 
ciples is very fully described by Haupt 
on pages 169 et seq. of his treatise. It 
is a wooden bridge over the Susquehanna 
River, 5^ miles from Harrisburg on the 

* Theory of Bridge Construction. Herman Haupt, A.M. 
New York. 1853. 



IN GRAPHICAL STATICS. 



45 



Pennsylvania Railroad, and was-designed 
by Haupt. It consists of twenty-three 
spans of 160 feet each from center to 
center of piers. The arches have each 
a span of 149:^ feet and a rise of 20 
ft. 10 in., and are stiffened by a Plowe 
Truss which is continuous over the 
piers and fastened to them. It was 
erected in 1849. Those parts which were 
protected from the weather have re- 
mained intact, while other parts have 
been replaced, as often as they have de- 
cayed, by pieces of the original dimen- 
sions. This bridge, though not designed 
for the heavy traffic of these days, still 
stands after twenty-eight years of use, a 
proof of the real value of this kind of 
combination in bridge building. 

CHAPTER XIII. 

THE ARCH OF MASONRY. 

Arches of stone and brick have joints 
which are stiff up to a certain limit 
beyond which they are unstable. The 
loading and shape of the arch must be so 
adjusted to each other that this limit 
shall not be exceeded. This will appear 
in the course of the ensuing discussion. 



Let us take for discussion the brick 
arch erected by Brunei near Maidenhead 
England, to serve as a railway viaduct. 

It is in the form of an elliptic ring, as 
represented in Fig. 14, having a span of 
128 ft. with a rise of 24 J feet. The 
thickness of the ring at the crown is 5j 
ft., while at the pier the horizontal thick- 
ness is 7 ft. 2 inches. 

Divide the span into an even number 
of equal parts of the type hh^ and with a 
radius of half the span describe the 
semicircle gg. Let 5a=24j ft. be the 
rise of the intrados, and from any con- 
venient point on the line hh as h^ draw 
lines to a and g. These lines will enable 
us to find the ordinates ha of the ellipse 
of the intrados from the ordinates hg of 
the circle, by decreasing the latter in the 
ratio of hg to ha. For example, draw a 
horizontal through g^ cutting h^g at i^^ 
then a vertical through ^3, cutting h^a at 
7*3, then will a horizontal through /^ cut 
off af)^ the ordinate of the ellipse corre- 
sponding to h^g^ in the circle, as appears 
from known properties of the ellipse. 

Similarly let ^$'=64 ft. + 7 ft. 2 in., 
and with hq as radius describe a semicir- 
cle. Let hd—1i\ix,. + 



h\ ft. be the rise 




Fig.14 
ARCH OF MASONRY 

MAIDENHEAD HAIL WAY VIADUCT 



46 



NEW CONSTRUCTIONS 



of the extrados, and from any convenient 
point on hh, as b^ draw lines to d and q. 
These will enable us to find the ordinates 
hd of the ellipse of the extrados, from 
those of the circle, by decreasing the 
latter in the ratio of hq to hd. By this 
means, as many points as may be desired, 
can be found upon the intrados and ex- 
trados; and these curves may then be 
drawn with a curved ruler. We can use 
the arch ring so obtained for our con- 
struction, or multiply the ordinates by 
any convenient number, in case the arch 
is too flat for convenient work. Indeed 
we can use the semicircular ring itself if 
desirable. We shall in this construction 
employ the arch ring ad which has just 
been obtained. 

We shall suppose that the material of 
the surcharge between the extrados and 
a horizontal line tangent at d causes by 
its weight a vertical pressure upon the 
arch. That this assumption is nearly 
correct in case this part of the masonry is 
made in the usual manner, cannot well be 
doubted. Rankine, however, in his Ap- 
plied Mechanics assumes that the press- 
ures are of an amount and in a direction 
due to the conjugate stresses of an homo- 
geneous, elastic material, or of a material 
which like earth has an angle of slope due 
to internal friction. While this is a cor- 
rect assumption, in case of the arch of a 
tunnel sustaining earth, it is incorrect 
for the case in hand, for the masonry of 
the surcharge needs only a vertical resist- 
ance to support it, and will of itself pro- 
duce no active thrust, having a horizon- 
tal component. 

This is further evident from Moseley's 
principle of least resistance, which is 
stated and proved by Rankine in the 
following terms: 

"If the forces which balance each 
other in or upon a given body or struc- 
ture, be distinguished into two systems, 
called respectively, active and passive^ 
which stand to each other in the rela- 
tion of cause and effect, then will the 
passive forces be the least which are 
capable of balancing the active forces, 
consistently with the physical condition 
of the body or structure. 

For the passive forces being caused by 
the application of the active forces to 
the body or structure, will not increase 
after the active forces have been balanced 



by them; and will, therefore, not increase 
beyond the least amount capable of bal- 
ancing the active forces." 

A surcharge of masonry can be sus- 
tained by vertical resistance alone, and 
therefore will exert of itself a pressure 
in no other direction upon the haunches 
of the arch. Nevertheless this surcharge 
will afford a resistance to horizontal 
pressure if produced by the arch itself. 
So that when we assume the prestures 
due to the surcharge to be vertical alone, 
we are assuming that the arch does not 
avail itself of one element of stability 
which may possibly be employed, but 
which the engineer will hesitate to rely 
upon, by reason of the inferior character 
of the masonry usually found in the sur- 
charge. The difficulty is usually avoided, 
as in that beautiful structure, the London 
Bridge, by forming a reversed arch over 
the piers which can exert any needed 
horizontal pressure upon the haunches. 
This in effect increases by so much the 
thickness of the arch ring at and near 
the piers. 

The pressure of earth will be treated 
in connection with the construction for 
the Retaining Wall. On combining the 
pressures there obtained with the weight, 
the load which a tunnel arch sustains, 
may be at once found, after which the 
equilibrium polygon may be drawn and 
a construction executed, similar in its 
general features to that about to be em- 
ployed in the case before us. 

Let us assume that the arch is loaded 
with a live load extending over the left 
half of the span, and having an intensity 
which when reduced to masonry of the 
same specific gravity as that of which 
the viaduct is built, would add a depth 
(?/* to the surcharge. Now if the number 
of parts into which the span is divided 
be considerable, the weights which may 
be supposed to be concentrated at the 
points of division vary very approximately 
as the quantities of the type af. This 
approximation will be found to be suffi- 
ciently exact for ordinary cases; but 
should it be desired to make the con- 
struction exact, and also to take account 
of the effect of the obliquity of the joints 
in the arch ring, the reader will find the 
method for obtaining the centers of 
gravity, and constructing the weights, in 
Woodbury's Treatise on the Stability of 
the Arch pp. 405 et seq. in which is 



IlSr GRAPHICAL STATICS. 



47 



given Poncelet's graphical solution of 
the arch. 

With any convenient pole distance, as 
one half the span, lay off the weights. 
We have used b as the pole and made 
h^w^ — i the . weight at the crown = 
J (af+ad) = b/tv^\ w^w,^ = a^f^, w^w^ = 
C6^f„, etc. Several of the weights near 
the ends of the span are omitted in the 
Figure; viz., w^w^, etc. From the force 
polygon so obtained, draw the equili- 
brium polygon c as previously explained. 

The equilibrium polygon which ex- 
presses the real relations between the 
loading and the thrust along the arch, is 
evidently one whose ordinates are pro- 
portional to the ordinates of the polygon 
c. 

It has been shown by Rankine, Wood- 
bury and others, that for perfect stability, 
— ^.e, in case no joint of the arch begins 
to open, and every joint bears over its 
entire surface, — that the point of appli- 
cation of the resultant pressure must 
everywhere fall within the middle third 
of the arch ring. For if at any joint the 
pressure reaches the limit zero, at the 
intrados or extrados, and uniformly in- 
creases to the edge farthest from that, 
the resultant pressure is applied at one 
third of the depth of the joint from the 
farther edge. 

The locus of this point of application 
of the resultant pressure has been called 
the " curve of pressure," and is evidently 
the equilibrium curve due to the weights 
and to the actual thrust in the arch. If 
then it be possible to use such a pole dis- 
tance, and such a position of the pole, 
that the equilibrium polygon can be in- 
scribed within the inner third of the 
thickness of the arch ring, the arch is 
stable. It may readily occur that this is 
impossible, but in order to ensure suffi- 
cient stability, no distribution of live 
load should be possible, in which this 
condition is not fulfilled. 

We can assume any three points at 
will, within this inner third, and cause a 
projection of the polygon c to pass 
through them, and then determine by in- 
spection whether the entire projection 
lies within the prescribed limits. In 
order to so assume the points that a new 
trial may most likely be unnecessary, we 
take note of the well known fact, that 
in arches of this character, the curve of 
pressure is likely to fall without the pre- 



scribed limits near the crown and near 
the haunches. Let us assume e at the 
middle of the crown, e/ at the middle of 
a/d/. and e^near the lower limit on a,d^, 

o o / 5 5 

This last is taken near the lower limit, 
because the curvature of the left half of 
the polygon is more considerable than 
the other, and so at some point between 
it and the crown it may possibly rise to 
the upper limit. The same consideration 
would have induced us to raise e/ to the 
upper limit, were it not likely that such 
a procedure would cause the polygon to 
rise above the upper limit on the right 
of e/. 

Draw the closing line kk through e^e/, 
and the corresponding closing line hh 
through c^c/, and decrease all the ordi- 
nates of the type he in the ratio of hb to 
ke, by help of the lines bn and bl, in a 
manner like that previously explained. 
For example h^c^=n^o^, and l^o^=k^e^. 
By this means we obtain the polygon e 
which is found to lie within the required 
limits. The arch is then stable: but is 
the polygon e the actual curve of 
pressures? Might not a different as- 
sumption respecting the three points 
through which it is to pass lead to a dif- 
ferent polygon, which would also lie 
within the limits ? It certainly might. 
Which of all the possible curves of pres- 
sure fulfilling the required condition, is 
to be chosen, is determined by Moseley's 
principle of least resistance, which ap- 
plied to the case in hand, would oblige 
us to choose that curve of all those lying 
within the required limits, which has the 
least horizontal thrust, z.6. the smallest 
pole distance. It appears necessary to 
direct particular attention to this, as a 
recent publication on this subject asserts 
that the true pressure line is that which 
approaches nearest to the middle of the 
arch ring, so that the pressure on the 
most compressed joint edge is a mini- 
mum; a statement at variance with the 
theorem of least resistance as proved by 
Rankine. 

Now to find the particular curve which 
has the least pole distance, it is evidently 
necessary that the curve should have its 
ordinates as large as possible. This may 
be accomplished very exactly, thus: 
above e^ where the polygon approaches 
the upper limit more closely than at any 
other point near the crown, assume a new 
position of e^ at the upper limit; and be- 



48 



TTEW CONSTRUCTIONS 



low e' where it approaches the lower 
limit most nearly on the right, assume a 
new position of e/ at the lower limit. 
At the left e^ may be retained. Now on 
passing the polygon through these points 
it will fulfill the second condition, which 
is imposed by the principle of least resist- 
ance. 

A more direct method for making the 
polygon fulfill the required condition 
will be given in Fig. 1 8. 

It is seen in the case before us, the 
changes are so minute that it is useless 
to find this new position of the polygon, 
and its horizontal thrust. The thrust ob- 
tained from the polygon e in its present 
position is suflBciently exact. The hori- 
zontal thrust in this case is found from 
the lines hn and hi. Since ""Ivv^ is the 
horizontal thrust, i.e. pole distance of the 
polygon c, ^vv^ is the horizontal thrust 
of the polygon e. 

By using this pole distance and a pok 
properly placed, we might have drawn 
the polygon e with perhaps greater ac- 
curacy than by the process employed, 
but that being the process employed in 
Figs. 2, 3, etc., we have given this as an 
example of another process. 

The joints in the arch ring should be 
approximately perpendicular to the 
direction of the pressure, i.e. normal to 
the curve of pressures. 

With regard to what factor of safety 
is proper in structures of this kind, all 
engineers would agree that the material 
at the most exposed edge should never 
be subjected to a pressure greater than 
one fifth of its ultimate strength. Owing 
to the manner in which the pressure is as- 
sumed to be distributed in those joints 
where the point of application of the re- 
sultant is at one third the depth of the 
joint from the edge, its intensity at this 
edge is double the average intensity of 
the pressure over the entire joint. We 
are then led to the following conclusion, 
that the total horizontal thrust (or pres- 
sure on any joint) when divided by the 
area of the joint where this pressure is 
sustained ought to give a quotient at 
least ten times the ultimate strength of 
the material. The brick viaduct which 
we have treated is remarkable in using 
perhaps the smallest factor of safety in 
any known structure of this class, having 



at the most exposed edge a factor of only 
3^ instead of 5. 

It may be desirable in a case like that 
under consideration, to discuss the 
changes occuring during the movement 
of the live load, and that this may be 
effected more readily, it is convenient to 
draw the equilibrium polygons due to 
the live and dead loads separately. The 
latter can be drawn once for all, while 
the former being due to a uniformly 
distributed load can be obtained with 
facility for different positions of the load. 
The polygon can be at once combined 
into a single polygon by adding the ordi- 
nates of the two together. Care must 
be taken, however, to add together only 
such as have the same pole distance. In 
case the construction which has been 
given should show that the arch is un- 
stable, having no projection of the equili- 
brium polygon which can be inscribed 
within the middle third of the arch ring, 
it is possible either to change the shape 
of the arch slightly, or increase its 
thickness, or change the distribution of 
the loading. The last alternative is 
usually the best one, for the shape has 
been chosen from reasons of utility and 
taste, and the thickness from considera- 
tion of the factor of safety. If the cen- 
ter line of the arch ring (or any other, 
line inscribed within the middle third) 
be considered to be an equilibrium poly- 
gon, and from a pole, lines be drawn 
parallel to the segments of this polygon, 
a weight line can be found which will 
represent the loading needed to make 
the arch stable. If this load line be 
compared with that previously obtained, 
it will be readily seen where a slight 
additional load must be placed, or else a 
hollow place made in the surcharge, 
such as will render the arch stable. In 
general, it may be remarked, that an 
additional load renders the curvature of 
the line of pressures sharper under it, 
while the removal of any load renders 
the curve straighter under it. 

The foregoing construction is unre- 
stricted, and applies to all unsymmetrical 
forms of arches or of loading, or both. 
As previously mentioned, a similar con- 
struction applies to the case of an arch 
sustaining the pressure of water or earth; 
in that case, however, the load is not ap- 
plied vertically and the weight line be- 
comes a polygon. 



IN GRAPHICAL STATICS. 



49 



CHAPTER XIV. 

RETAINING WALLS AND ABUTMENTS. 

Let aa'h'h in Fig. 15 represent the 
cross section of a wall of masonry which 
retains a bank of earth having a surface 
aa^. Assume that the portion of the 
wall and earth under consideration is 
bounded by two planes parallel to the 
plane of the paper, and at a unit's dis- 
tance from each other: then any plane 
containing the edge of the wall at ^, as 
ha^^ ha^^ etc., cuts this solid in a longitu- 
dinal section, which is a rectangle having 
a width of one unit, and a length ha^^ ba^, 
etc. 

The resultant of the total pressure 
distributed over any one of these rec- 
tangles of the type ha is applied at one- 
third of that distance from b\ i.e. the re- 
sultant pressure exerted by the earth 
against the rectangle at ha^ is applied at 
a distance of M'=^ ha^ from b. 

That the resultant is to be applied at 
this point, is due to the fact that the dis- 
tributed pressure increases uniformly as 



we proceed from any point a of the sur- 
face toward b: the center of pressure is 
then at the point stated, as is well known. 

Again, the direction of the pressures 
against any vertical j^lane, as that at ba^^ 
is parallel to the surface aa.^. This fact 
is usually overlooked by those who treat 
this subject, and some arbitrary assump- 
tion is made as to the direction of the 
pressure. 

That the thrust of the earth against 
a vertical plane is parallel to the ground 
surface is proved analytically in Ran- 
kine's Applied Mechanics on page 127; 
which proof may be set forth in an 
elementary manner by considering the 
small parallelepiped mn, whose upper 
and lower surfaces are parallel to the 
ground surface. Since the pressure on 
any plane parallel to the surface of the 
ground is due to the weight of the earth 
above it, the pressure on such a plane is 
vertical and uniformly distributed. If 
mn were a rigid body, it would be held 
in equilibrium by these vertical pressures, 
which are, therefore, a system of forces 




Fig.l5 
THRUST OF EARTH 

DETAINING WALL 



50 



NEW CONSTRUCTIOIvrS 



in equilibrium; but as tnn is not rigid it 
must be confined by pressures distributed 
over each end surface, which last are dis- 
tributed in the same manner on each end, 
because each is at the same depth below 
the surface. I^ow the vertical pressures 
and end pressures hold 7nn in equilibrium^ 
they therefore form a system in equili- 
brium. But the vertical pressures are in- 
dependently in equilibrium, therefore the 
end pressures alone form a system which 
is independently in equilibrium. That this 
may occur, and no couple be introduced, 
these must directly oppose each other; 
i.e. be parallel to the ground line aa^. 

Draw kp \\ aa^, it then represents the 
position and direction of the resultant 
pressure upon the vertical ha^. Draw 
the horizontal H, then is the angle ik2:> 
called the ohliqidty of the pressure, it 
being the angle between the direction of 
the pressure and the normal to the plane 
upon which the pressure acts. 

Let ehc=-^hQ the angle of friction, i.e. 
the inclination which the surface of 
ground would assume if the wall were 
removed. 

The obliquity of the pressure exerted 
by the earth against any assumed plane, 
such as ha^ or ba^, must not exceed the 
angle of friction; for should a greater 
obliquity occur the prism of earth, aj)a^ 
or afia^, would slide down the plane, ba^ 
or ba^, on which such obliquity is found. 

For dry earth ^ is usually about 30°; 
for moist earth and especially moist clay, 
^ may be as small as 15°. The inclina- 
tion of the ground surface aa^ cannot be 
greater than ^. 

Now let the points a^, a^, a^, etc., be 
assumed at any convenient distances 
along the surface: for convenience we 
have taken them at equal distances, but 
this is not essential. With ^ as a center 
and any convenient radius, as be, describe 
a semi-circumference cutting the lines 
ha^, ba^, etc. at c^, c^, etc. Make ee^-=ec; 
also e/^^CgCj, e^e^^c^c^, etc.: then be^ 
has an obliquity ^ with ba^, as has also 
he^ with ba^, ^e, " " 
= afie^ — aJ^e^-^^^°-\- ^. 

Lay off bb^, bb^, bb^, etc., proportional 
to the weights of the prisms of earth 
a})a^, ctfia^, a^ba^, etc.: we have effected 
this most easily by making a^a^ = bb^, 
a^a^ — bb^, a^a^ — bb^, etc. Through J, b^ b^, 
etc., draw parallels to kp\ these will inter- 
sect be^, be^, be^, etc., at b, t^, t^, etc. 



with ba^y etc.; for a^be^ 



Then is bb^t^ the triangle of forces hold- 
ing the prism a^ba^ in equilibrium, just 
as it is about to slide down the plane ba^, 
for. bb^ represents the weight of the 
prism, b^tj^ is the known direction of the 
thrust against ba^, and bt^ is the direc- 
tion of the thrust against ba^ when it is 
just on the point of sliding: then is t^b^ 
the greatest pressure which the prism 
can exert agsCinst ba^. Similarly t^b^ is 
the greatest pressure which the prism 
afia^ can exert. ISTow draw the curve 
t^t^t^, etc., and a vertical tangent inter- 
secting the parallel to the surface through 
b a,t t; then is tb the greatest pressure 
which the earth can exert against ba^. 
This greatest pressure is exerted approxi- 
mately by the prism or wedge of earth 
cut off by the plane ba^, for the pressure 
which it exerts against the vertical plane 
through b is almost exactly bj>^=bt. 
This is Coulomb's " wedge of maximum 
thrust" correctly obtained: previous de- 
terminations of it have been erroneous 
when the ground surface was not level, 
for in that case the direction of the press- 
ure has not been ordinarily assumed to 
be parallel to the ground surface. 

In case the ground surface is level the 
wedge of maximum thrust will always 
be cut off by a plane bisecting the angle 
cbc^, as maybe shown analytically, which 
fact will simplify the construction of that 
case, and enable us to dispense with 
drawing the thrust curve U. 

The pressure tb is to be applied at k, 
and may tend either to overturn the wall 
or to cause it to slide. 

In order to discuss the stability of the 
wall under this pressure, let us find the 
weight of the wall and of the prism of 
earth aba^. Let us assume that the 
specific gravity of the masonry compos- 
ing the wall is twice that of earth. 
Make a^h=bb', then the area abb'a'^ 
abli=-abh^\ and if ah^^^2ah, then ah^ 
represents the weight of the wall reduced 
to the same scale as the prisms of earth 
before used. Since aa^ is the weight of 
cibci^^ aji^ is the weight of the mass on 



-05 



the right of the vertical ba^ against 
which the pressure is exerted. 

Make bq=-aji^, and draw tq, which 
then represents the direction and amount 
of the resultant to be applied at o where 
the resultant pressure applied at k inter- 
sects the vertical gw through the center 
of gravity g of the mass aajbb'a' . The 



IN GEAPHICAL STATICS. 



51 



center of gravity g is constructed in the 
following manner. Lay off a'h^-W ^ and 
hl=^aa' \ and join hi. Join also the mid- 
dle points of ah and a'b': the line so 
drawn intersects Jil at g^ the center of 
gravity of aa'h'h. Find also the center 
of gravity g„_^ of ciba^^ which lies at the 
intersection of a line parallel to aa^, and 
cutting ha^ at a distance of J ba^ from a^ 
and of a line from h bisecting aa^. 
Through g^ and g^ draw parallels, and 
lay off g.J\ and^,/!^ on them proportional 
to the weights applied at g^ and g^ 
respectively. We have found it con- 
venient to make g^f ^=^i(^(^h^, and g^f^=:^ 
aa^. Then/'j/*2 divides g^g^ inversely as 
the applied weights; and g, the point of 
intersection, is the required center of 
gravity. 

Let or be parallel to tq ; since it 
intersects bb' so far within the base, 
the wall has sufficient stability against 
overturning. The base of the wall is so 
much greater than is necessary for the 
support of the weight resting upon it, 
that engineers have not found it neces- 
sary that the resultant pressure should 
intersect the base within the middle third 
of the joint. The practice of English en- 
gineers, as stated by Rankine, is to per- 
mit this intersection to approach as near 
b' as \bb', while French engineers permit 
it to approach as near as \bh' only. In 
all cases of buttresses, piers, chimneys, 
or other structures which call into play 
some fraction of the ultimate strength 
of the material, or ultimate resistance of 
the foundation as great as one tenth, or 
one fifteenth, the point should not ap- 
proach b' nearer than ^ bb'. 

Again, let the angle of friction be- 
tween the wall and the earth under it be 
^' : then in order that the thrust at k 
may not cause the wall to slide, the 
angle wor must be less than ^'. 

When, however, the angle ^' is less than 
wor itbecomes necessary to gain additional 
stability by some means, as for example 
by continuing the wall below the sur- 
face of the ground lying in front of it. 
Let «/«/ be the surface of the ground 
which is to afford a passive resistance to 
the thrust of the wall: then in a manner 
precisely analogous to that just employed 
for finding the greatest active pressure 
which earth can exert against a vertical 
plane, we now find the least passive 
pressure which the earth in front of the ! 



wall will sustain without sliding up some 
plane such as b'a^' or b'a/, etc. The 
difference in the two cases is that in the 
former case friction hindered the earth 
from sliding down, while it now hinders 
it from sliding up the plane on which it 
rests. * 

Lay off e'e/=6e„; then taking any 
points a^'a^', etc. on the ground surface, 
make e./ej^=cjcj^ eje/=^cjcj^ etc. 



'1 J 



'3 5 



Lay off b'b^'-=a^'a^\ etc., and drawing 
parallels through ^/, b^\ etc., we obtain 
the thrust curve t^'t^\ etc. 

The small prism of earth between b'aj 
and the wall adds to the stability of the* 
wall, and can be made to enter the con- 
struction if desired, in the same manner 
as did aba^. 

The vertical tangent through s' shows 
us that the earth in front of the wall can 
withstand a thrust having a horizontal 
component b's' measured on a scale such 
that b'b^^=^a^a^ is the weight of the 
prism of earth ajb'a^. 

This scale is different from that used 
on the left. To reduce them to the 
same scale lay off from b\ the distances 
b'd^ and b'dj proportional to the perpen- 
diculars from b on aa, and b^ on a/ a' 

4 14 

respectively. In the case before us, as 
the ground surfaces are parallel, we have 
made b'd^-=^ba^ and b'd^=.b'a^. ' 

Then from any convenient point on 
b'b^., as V, draw vd^ and vd^\ these lines 
will reduce from one scale to the other. 
We find then that ic'f? is the thrust on 
the scale at the left corresponding to 
xd^b's' on the right: i.e., the earth 
under the surface assumed at the right 
can withstand something over one fourth 
of the thrust sb at the left. 

It will be found that a certain small 
portion of the earth near a/ has a thrust 
curve on the left of b', but as it is not 
needed in our solution it is omitted. 

If any pressure is required in pounds, 
as for example sb, it is founds as follows: 
— the length of ah^ is to that of sb as the 
weight of bb'aa' in lbs. is to the pressure 
S'5 in lbs. 

Frequently the ground surface is not a 
plane, and when this is the case it often 
consists of two planes as ad, da^ Fig. 16. 
In that case, draw some convenient line 
as ad^, and lay off ad^, d^d.^, etc. at will, 
which for convenience we have made 
equal. Draw d^a^, <^^2^25 ®*^* parallel to 
bd, and join ba^, ba^, etc.: then are the 



52 



NEW CONSTRUCTIONS 



^ 



0-7 <^5 "5 rt4 rtsd 



\\ \\^v^ 



Fig.16 




ado ci 




triangles bda, bcla^, bda^, bda^, etc. pro- 
portional in area to the lines ea, ea^, etc. 
Hence the weights of the prisms of earth 
baa^y baa^, etc., are proportional to ad^, 
ad^, etc. 

In case ab slopes backward the part of 
the wall at the left of the vertical ba^ 
rests upon the earth below it sufficiently 
to produce the same pressure which 
would be produced if baa^ were a prism 
of earth. The weights of the wedges 
which produce pressures, and which are 
to be laid off below b, are then propor- 
tional to d^d^=bb^, d^d^ = bb^, etc. The 
direction of the pressures of the prisms 
at the right of bd are parallel to ad; but 
upon taking a larger prism the direction 
may be assumed to be parallel to a^a^, 
ci^a^y etc., which is very approximately 
correct. Now draw b^t^ || a„a^, b^t^ \\ a^a^, 
etc.; and complete the construction for 
pressure precisely as in Fig. 15, using 
for resultant pressure the direction and 
amount of that due to the wedge of maxi- 
mum pressure thus obtained. 

In finding the stability of the wall, it 
will be necessary to find the weight and 
center of gravity of the wall itself, minus 
a prism of earth baa^, instead of plus this 
prism as in Fig. 15; for it is now sus- 
tained by the earth back of the wall. 

When the back of the wall has any 



other form than that above treated, the 
vertical plane against which the pressure 
is determined should still pass through 
the lower back edge of the wall. 

In case the wall is found to be likely 
to slide upon its foundations when these 
are level, a sloping foundation is fre- 
quently employed, such that it shall be 
nearly perpendicular to the resultant pres- 
sure upon the base of the wall. The con- 
struction employed in Fig. 15 applies 
equally to this case. 

The investigation of the stability of 
any abutment, buttress, or pier, against 
overturning and against sliding, is the 
same as that of the retaining wall in Fig. 
15. As soon as the amount, direction, 
and point of application, of the pressure 
exerted against such a structure is deter- 
mined, it is to be treated precisely as 
was the resultant pressure kp in Fig. 15. 

In the case of a reservoir wall or dam, 
the construction is simplified from the 
fact ^hat, since the surface of water is 
level and the angle of friction vanishes, 
the resultant pressure is perpendicular 
to the surface upon which the water 
presses. It is useful to examine this as 
a case of our previous construction. In 
Fig. 17, let abb^ be the cross-section of 
the dam; then the wedge of maximum 
pressure against ba^ is cut off by the 



IN GRAPHICAL STATICS. 



53 



plane ha^ when cJaj=45°, i.e. ha^ bisects 
cba^ as before stated. 




This produces a horizontal resultant 
pressure at k equal to the weight of the 
wedge. Now the total pressure on ah is 
the resultant of this pressure, and the 
weight of the wedge aha^. The forces 
to be compounded are then proportional 
to the lines a^a^ = bv^ and aa^. By simi- 
larity of triangles it is seen that ro the 
resultant is perpendicular to ab. 

It is seen that by making the inclina- 
tion of ab small, the direction of ro can 
be made so nearly vertical that the dam 
will be retained in place by the pressure 
of the water alone, even though the dam 
be a wooden frame, whose weight maybe 
disregarded. 

We can now construct the actual 
pressures to which the arch of a tunnel 
surcharged with water or earth is sub- 
jected. Suppose, for example, we wish 
to find the pressure of such a surcharge 
on the voussoir a^d^d^a^ Fig. 14. Find 
the resultant pressure against a vertical 
plane extending from d^ to the upper 
surface of the surface and call it 2^^- 
Draw a horizontal through d^ and 
let its intersection with the vertical 
just mentioned he called d". Find 
the resultant pressure against the verti- 
cal plane extending from d" to the sur- 
face, and call it p\ ISTow let p" ^=- 
^:).— ^/and let it be applied at such a point 
of d^d" that p.^ shall be the resultant of 79/ 
and p" . Then will the resultant press- 
ure against the voussoir be the resultant 
oi p^ and the weight of that part of the 
surcharge directly above it. 

FOUNDATIONS IN EARTH. 

A method similar to that employed in 
the determination of the pressure of 
«arth against a retaining wall, or a tunnel 
arch, enables us to investigate the sta- 
bility of the foundations of a wall stand- 
ing in earth. 

Suppose in Fig. 15 that the wall abb' a' 
is a foundation wall, and that the press- 
tire which it exerts upon the plane bb' 
is vertical, being due to its own weight 
.and the weio'ht of the building or other 



load which it sustains. Now consider a 
vertical plane of one unit in height, say, 
as hb^ ; and determine the resultant press- - 
ure against it on the supposition that 
the pressure is produced by a depth of 
earth at the right of it, sufficient to pro- 
duce the same vertical pressure on bh' 
which the wall and its load do actually 
produce. In other words we^suj^pose 
the wall and load replaced by a bank of 
earth having its upper surface horizontal 
and weighing the same as the wall and 
load. Call the upper surface z^ and find 
the pressure against the vertical plane zb 
due to the earth under the given level 
surface; similarly, find the pressure 
against zh^. The surface being level, the 
maximum pressure, as previously stated 
will be due to a wedge cut off by a plane 
bisecting the . angle between bz and a 
plane drawn from b at the inclinatian ^, 
of the limiting angle of friction. This 
enables us to find the horizontal pres- 
sures against zb and zb^ directly: their 
difference is the resultant active pressure 
against bb^. 

Next, it must be determined what pas- 
sive pressure the earth at the left of bb^ 
can support. The passive resistance of 
the earth under the surface a ao^ainst 
the plane ab as well as that against the 
plane ab^ can be found exactly as that 
was previously found under the surface 
a' . The difference of these resistances is 
the resistance which it is possible for bb^ 
to supj^ort. Indeed bb^ could support 
this pressure and afford this resistance 
even if the active pressure against ab 
were, at the limit of* its resistance, which 
it is not. The limiting resistance which 
is thus obtained, is then so far within 
the limits of stability, that ordinarily, no 
further factor of safety is needed, and 
the stability of the foundation is secured, 
if the active pressure against bb^ does not 
exceed the passive resistance. This con- 
struction should be made on the basis of 
the smallest angle of friction ^ which 
the earth assumes when wet; that being 
smaller than for dry earth, and hence 
giving a greater active pressure at the 
right, and a less resistance at the left. 

CHAPTER XV. 

SPHERICAL DOME OF METAL. 

The dome which will be treated in the 
following construction is hemispherical 
in shape; but the proposed construction 



54 



NEW CONSTKUCTIONS 



applies equally to domes of any different 
form generated by the revolution of the 
arc of some curve about a vertical axis : 
such forms are elliptic, parabolic or hy- 
perbolic domes, as well as pointed or 
gothic domes, etc. Let the quadrant aa 
in Fig. 18, represent the part of the 
meridian section of a thin metallic dome 
between the crown and the springing 
circle. The metallic dome is supposed 
to be so thin that its thickness need not 
be represented in the Figure : the thick- 
ness of a dome of masonry, however, is a 
matter of prime importance and will be 
treated subsequently. 

In a thin metallic dome the only thrust 
along a meridian section is necessarily 
in a direction tangent to that section at 
each point of it. This consideration will 
enable us to determine this thrust as well 



as the hoop tension or compression along 
any of the conical rings into which the 
dome may be supposed to be divided 
by a series of horizontal planes. 

Let the height ah of .the dome be 
divided into any number of parts, which 
we have in this case, for convenience, 
made equal. Let these equal parts of the 
type c?w be the distances between horizon- 
tal planes such that the planes through 
thepoints.c?,, c?^, etc., cut small circles from 
the hemisphere which pass through the 
point a^, ^2) Gtc, and similarly the planes 
through Wj, ^2, etc., cut small circles which 
pass through g^^ g^^ etc. Now suppose the 
thickness of this dome to be uniform, 
and if ah be taken to represent the weight 
of a quadrantal lune of the dome included 
between two meridian planes making 
some small angle with each other; then 



SPHERICAL DOME 

rig.is 




IN GRAPHICAL STATICS. 



55 



from the well-known expression for the 
area of the zone of a sphere it appears that 
ad^ will represent the weight of that 
part of the lune above a^d^. Similarly 
an^ is the weight of the lune • ag^ ; 
ad^ the weight of aa^^ etc. 

This method of obtaining the weight 
applies of course in case the dome is any 
segment of a sphere less than a hemi- 
sphere and of uniform thickness. If the 
thickness increases from the crown, the 
weights of the zones cut by equi-distant 
horizontal planes increase directly as the 
thickness. In case the dome is not 
spherical the weights must be determin- 
ed by some process suited to the form of 
the dome and its variation in thickness. 

Now the weight of the lune aa^ is sus- 
tained by a horizontal thrust which is 
the resultant of the horizontal pressures 
in the meridian planes by which it is 
bounded, and by a thrust, as before re- 
marked, in the direction of the tangent 
at a* Draw a horizontal line through d^^ 
and through a a parallel to the tangent 
at a: these intersect at s^, then is ad^s^^ 
the triangle of forces which hold in 
equilibrium the lune aa^. Similarly, 
au^t^ is the triangle of forces holding the 
lune ag^ in equilibrium, etc. Draw a 
curve St through the points thus determ- 
ined. This curve is a well-known cubic 
which when referred to ba as the axis of 
X and bg^ as that of y has for its equa- 
tion 



x' 



r—x 

r + x 



On being traced at the right of a it has 
in the other quadrant of the dome a part 
like that here drawn forming a loop; it 
passes through b at an inclination of 45° 
and the two branches below b finally 
become tangent to a horizontal line 
drawn tangent to the circle aa of the 
dome. The curve has this remarkable 
property : — If any line be drawn from a, 
cutting the curve here drawn and, also, 
the part below bg^, the product of these 
two radii vectores of the curve from the 
pole a is constant, and the locus of the 
intersection of the normals at these two 
points is a parabola. 

Draw a vertical tangent to this curve : 
the point of contact is very near ^3, and g^, 
the corresponding point of the dome is 
almost 52° from the crown a. A determi- 
nation of this maximum point by means 



of the equation gives the height of it 
above b sls ^ (a/5~-^) ^'» corresponding to 
about51°49'. Now consider any zone, as, 
for example, that whose meridian section 
is g/i^: the upper edge is subjected to a 
thrust whose radial horizontal compo- 
nent is proportional to ti^t^, while the 
horizontal thrust against its lower edge 
is proportional to d^s^, and the difference 
s^x^ between these radial forces produces 
a hoop compression around the zone pro- 
portional to s^x^. It will be seen that 
these differences which are of the type 
sx or tg, change sign at t^. Hence all 
parts of the dome above 51° 49' from the 
crown, are subjected to a hoop compres- 
sion which vanishes at that distance from 
a, while all parts of the dome below 
this are subjected to hoop tension. This 
may be stated by saying that a thin 
dome of masonry would be stable under 
hoop compression as far as 51^ 49' from 
the crown, but unstable below that, being 
liable to crack open along its meridian 
sections. A thick dome of masonry, 
however, does not have the resultant 
thrust at every point of its meridian 
section in a direction which is tangential 
to its surface, — this will be discussed 
later. 

It is necessary to determine the actual 
hoop tension or compression in any ring 
in order to determine the thickness of 
the dome such that the metal may not 
be subjected to too severe a stress. 

The rule for obtaining hoop tension 
(we shall use the word tension to in- 
clude both tension and compression) is : 
Multiply the intensity of the radial 
pressure by the radius of the hoop, the 
product is the tension at any meridian 
section of the hoop. The correctness of 
this rule appears at once from considera- 
tion of fluid pressure in a tube, in which 
it is seen that the tensions at the two ex- 
tremities of a diameter prevent the total 
pressure on that diameter from tearing 
the tube asunder. 

Now in the case before us t^g^ is the 
radial force distributed along a certain 
lune. The number of degrees of which, 
the lune consists is at present undeterm- 
ined : let it be determined on the suppo- 
sition that it shall be such a number of 
degrees as to cause that the total radial 
force against it shall be equal to the 
hoop tension. Call the total radial force 
P and the hoop tension T, then the lune 



56 



NEW CONSTRUCTIONS 



is to be such that P= T. Also let Q be 
the number of degrees in the lune, then 
90° -4-^ is the number of lunes in a quarter 
of the dome, and 90 P-^O is the radial 
force against a quarter of the dome, 
which last must be divided by Jtt to ob- 
tain the hoop tension; because if ^:> is the 
intensity of radial pressure, ^nrj^ is the 
total pressure against a quadrant and rj), 
as previously stated, is the hoop tension. 
The ratio of these is ^tt, and by this we 
must divide the total radial pressure in 
every case to obtain hoop tension 



180 P _ 
' On ~ ' 

ior P=T 



6^ 



180^ 



7t 



This is the number of degrees of which 
the lune must consist in order that when 
ab represents its weight, t^y^ shall rep- 
resent the hoop tension in the meridian 
section ct^g^. The expression we have 
found is independent of the radius of the 
ring, and hence holds for any other ring 
as g^a^, in which s^x^ is the hoop tension, 
etc. To find what fraction this lune is 
of the whole dome, divide 6 by 360° 



e 



180 



nearly, 



360 360;r 27t 25 

from which the scale of weight is easily 
found, thus; let TK be the total weight 
of the dome and r its radius, then 

271)' : W\ ',1 : n, the weight per unit, or 
the hoop tension per unit of the distances 
ti/ or sx. 

Distances a^ or as, on the same scale, 
represent the thrust tangential to the 
dome in the direction of the meridian 
sections, and uniformly distributed over 
an arc of 5'7°.3— : e.g. if we divide at^ 
measured as a force by X u^g^ measured 
as a distance we shall obtain the intensi- 
ty of the meridian compression at the 
joint cut from the dome by the horizon- 
tal plane through a^. 

Analogous constructions hold for 
domes not spherical and not of uniform 
thickness. Approximate results may be 
obtained by assuming a spherical dome, 
or a series of spherical zones approxi- 
mating in shape to the form which it is 
desired to treat. 



CHAPTER XVI. 

SPHERICAL DOME OF MASO]S^RY. 

Let the dome treated be that in Fig. 
18 in which the uniform thickness of the 
masonry is one-sixteenth of the internal 
diameter or one-eighth of the radius of 
the in trad OS. Divide ab the radius of 
the center line into any convenient num- 
ber of equal parts, say eight, at tc^, u^, 
etc.: a much larger number would be 
preferable in actual construction. At 
the points a^, a^, etc., on the same levels 
with ^<^, 1^2, etc. pass conical joints nor- 
mal to the dome, so that b is the vertex 
of each of the cones. 

If we consider a lune between meridian 
planes making a small angle with each 
other, the center of gravity of the parts 
of the lune between the conical joints lie 
at g^, ^2) 6tc. on the horizontal midway 
between the previous horizontals. These 
points are not exactly upon the central 
line aa, but if the number of horizontals 
is large, the difference is inappreciable. 
We assume them upon aa. That they 
fall upon the horizontals through d^, d^^ 
etc., midway between those through u^, 
t/.^, etc., is a consequence of the equality 
in area between spherical zones of the 
same height. 

In finding the volume of a sphere it 
may be considered that, we take the sum 
of a series of elementary cones whose 
bases form the surface of the sphere, and 
whose height is the radius. Hence, if 
any equal portions of the surface of a 
sphere be taken and sectorial solids be 
formed on them as bases and having 
their vertices at the center, then the 
sectorial solids have equal volumes. 
The lunes of which we treat are equal 
fractions of such equal solids. 

Draw the verticals of the type bg 
through the centers of gravity g^, g^, etc. 
The weights applied at these points are 
equal and may be represented by au^, 
it^u^^^w^io^, etc. Use a as the pole and 
w^w^ as the weight line; and, beginning 
at the 23oint /"g, draw the equilibrium 
polygon c due to the weights. 

We have used for pole distance the 
greatest horizontal thrust which it is 
possible for any segment of the dome to 
exert upon the part below it, when the 
hoop compression extends to 51° 49' 
from the crown. 

Below the point where the compression 



IN GRAPHICAL STATICS. 



57 



vanishes we shall not assume that the 
bond of the masonry is such that it can 
resist the hoop tension which is develop- 
ed. The upper part of the dome will be 
then carried by the parts of the lunes 
below this point by their united action 
as a series of masonry arches standing 
side by side. 

Now it is seen that the curve of equi- 
librium c, drawn with this assumed hori- 
zontal thrust falls within the curve of the 
lune, which signifies that the dome will 
not exert so great a thrust as that as- 
sumed. By the principle of least resist- 
ance, no greater horizontal thrust will 
be called into action than is necessary to 
cause the dome to stand, if stability is 
possible. If a less thrust than that just 
employed be all that is developed in the 
dome, then the point where the hoop 
compression vanishes is not so far as 51° 
49' from the crown, and a longer portion 
of the lune acts as an arch, than has been 
supposed by previous writers on this 
subject,* none of whom, so far as known, 
have given a correct process for the solu- 
tion of the problem, although the results 
arrived at have been somewhat approxi- 
mately correct. 

To ensure stability, the equilibrium 
curve must be inscribed within the inner 
third of that part of the meridian section 
of the lune which is to act as an arch ; as 
appears from the same reasons which 
were stated in connection with arches of 
masonry. 

And, further, the hoop compression 
will vanish at that level of the dome 
where the equilibrium curve, in departing 
from the crown, first becomes more 
nearly vertical than the tangent of the 
meridian section; for above that point 
the greatest thrust that the dome can 
exert, cannot be so great as at this point 
where the thrust of the arch-lune is equal 
to that of the dome. 

Now to determine in what ratio the 
ordinates of the curve c must be elongat- 
ed to give those of the curve e which 
fulfills the required conditions, we draw 
the line /o, and cut it at ^9„ p^^ etc. by 
the horizontals m^p^^ '^iPit ^tc, the quan- 
tities rah being the ordinates of exterior 
of the inner third. Again draw verticals 
through /)j, ^2) ^tC'j ^"<^ ^^^ them at q^^ 

* See a paper read before the Royal Inst, of British 
Architects, "on the Mathematical Theory of Domes," 
Feb. 6th, 1S71. By Edmund Beckett Denison, L.L.D., 
Q.C., F.R.A.S. 



$'2' ?35 ^^^' ^y horizontals through c„ c^, 
C3, etc. Through these points draw the 
curve qq^ whose ordinates are of the type 
qh. Some one of these ordinates is to 
be elongated to its corresponding ph^ 
and in such a manner that no qh shall 
then become longer than its correspond- 
ing ph. To eifect this, draw oq^ tangent 
to the curve qq\ then will oq^ enable us 
to effect the required elongation: e.g. let 
the horizontal through c^ cut oq^ at y^, 
and then the vertical through j ^ cuts/b 
at i^, then is e^ (which is on the same 
level with i^ the new position of c^. 
Similarly, we may find the remaining 
points of the curve e; but it is better to 
determine the new pole distance, and use 
this method as a test only. 

The curve qq made use of in this con- 
struction for finding the ratio lines for 
so elongating the ordinates of the curve 
e, that the new ordinates shall be those 
of a curve e tangent to the exterior line 
of the inner third, may be applied with 
equal facility to the construction for the 
arch of masonry. This furnishes us with 
a direct method in place of the tentative 
one employed in connection with Fig. 
14. 

To find the new pole distance, draw 
fj II oq^ cutting ww at /, then will i the 
intersection of the horizontal through j^ 
be the new position of the weight line liV)^ 
having its pole distance from a diminish- 
ed in the required ratio. 

The equilibrium curve e will be parallel 
to the curve of the dome at the points 
where the new weight line vv cuts the 
curve 8t. It should be noticed that the 
pole distance which we have now determ- 
ined is still a little too large because 
the polygon e is circumscribed about 
the true equilibrium curve; and as the 
polygon has an angle in the limiting 
curve mvin the equilibrium curve is 
not yet high enough to be tangent to the 
limiting curve. If the number of divi- 
sions had originally been larger (which 
the size of our Figure did not permit) 
this matter would be rectified. 

The polygon e is seen at e^ to fall just 
without the required limits, this would 
be partly rectified by slightly decreasing 
the pole distance as just suggested; the 
point, however, would still remain just 
without the limit after the pole distance 
is decreased, and by so much is the dome 
unstable. A dome of which the thick- 



58 



NEW CONSTRUCTIONS 



ness is one fifteenth of the internal dia- 
meter, is almost exactly stable. 

It is a remarkable fact that a semi- 
cylindrical arch of uniform thickness and 
without surcharge must be almost exact- 
ly three times as thick, viz., the thickness 
must be about one fifth the span in order 
that it may be possible to inscribe the 
equilibrium curve within the inner third. 

The only large hemispherical dome, of 
which I have the dimensions, which is 
thick enough to be perfectly stable with- 
out extraneous aid such as hoops or ties, 
is the Gol Goomuz at Beejapore, India. 
It has an internal diameter of 137^ feet, 
and a thickness of 10 feet, it being 
slightly thicker than necessary, but it 
probably carries a load upon the crown 
which requires the additional thickness. 

The hemispherical dome of uniform 
thickness is a very faulty arrangement 
of material. It is only necessary to 
make the dome so light and thin for 51° 
49' from the crown that it cannot exert 
so great a horizontal thrust as do the 
thicker lunes below, to take complete ad- 
vantage of the real strength of this form 
of structure. A dome whose thickness 
gradually decreases toward the crown 
takes a partial advantage of this, but 
nothing short of a quite sudden change 
near this point appears to be completely 
effective. 

The necessary thickness to withstand 
the hoop compression and the meridian 
thrust can be found as previously shown 
in the dome of metal. 

Domes are usually crowned with a 
lantern or pinnacle, whose weight must 
be first laid off below the pole a after 
having been reduced to the same unit 
as that of the zones of the dome. 

Likewise when there is an eye, at the 
crown or below, the weight of the mate- 
rial necessary to fill the eye must be sub- 
tracted, so that a is then to be placed 
below its present position. The construc- 
tion is then to be completed in the same 
manner as in Fig, 18. 

It is at once seen that the effect of an 
additional weight, as of a lantern, at the 
crown, since it moves the point a upward 
a certain distance, will be to cause the 
curve St to have all its points except b to 
the left of their present position, and 
especially the points in the upper part of 
the curve, thus making the point of no 
hoop tension much nearer the crown than 



in the metallic dome. It will be noticed 
that the addition of very small weight at 
the crown will cause the point ni^ of no 
hoop tension in the dome of masonry to 
approach almost to the crown, so that 
then the lunes will act entirely as stone 
arches with the exception of a very small 
segment at the crown. 

On the contrary, the removal of a seg- 
ment at the crown, or the decrease of the 
thickness, or any device for making the 
upper part of the dome lighter will re- 
move the point of no hoop tension further 
from the crown, both for the dome of 
metal and of masonry. In any dome of 
masonry the thickness above the point 
of no hoop tension, as determined by the 
curve St, need be only such as to with- 
stand the two compressions to which it 
is subjected, viz; hoop compression and 
meridian compression: while below that 
the lunes acting as arches must be thick 
enough to cause a horizontal thrust equal 
to the maximum radial thrust of the 
dome above the point of no hoop ten- 
sion. 

Several large domes are constructed of 
more than one shell, to give increased 
security to the tall lanterns surmounting 
them : St. Peter's, at Rome, is double, 
and the Pantheon, at Paris, is triple. 
The different shells should all spring 
from the same thick zone below the 
point of no hoop tension; and the lunes 
of this thick zone should be able to 
afford a horizontal thrust equal to the 
sum of the radial thrusts of all the 
shells standing upon it. 

Attention to this will secure the sta- 
bility in itself of any dome of masonry 
spherical or otherwise; and, though I 
here offer no proof of the assertion, I am 
led to believe that this is the solution of 
the problem of constructing the dome of 
a minimum weight of material, on the 
supposition that the meridian joints can 
afford no resistance to hoop tension. 

Now, in fact, it is a common device to 
ensure the stability of large domes by 
encircling them with iron hoops or 
chains, or by embedding ties in the ma- 
sonry; and this case appears to be of 
sufficient importance to demand our at- 
tention. 

If the hoop encircles the dome at 51® 
49' or any other less distance from the 
crown the dome will be a true dome at 
all points above the hoop. Suppose the 



IN GRAPHICAL STATICS. 



59 



hoop to be at 51° 49', then the curve e 
should, below that point, be made to 
pass through the points f\ and /"g, from 
which it is seen that the dome may be 
made thinner than at present, and the 
horizontal thrust caused will be less. 
The tension of the hoop would be that 
due to a radial thrust which is the dif- 
ference between that given by the curve 
St for this point and the horizontal thrust 
(pole distance) of the polygon e when it 
passes through /g and/g. That the curve 
e passes through these last mentioned 
points is a consequence of the principle 
of least resistance. 

Again, suppose another hoop encircles 
the dome at/'^; the curve e must pass 
through ^/"g and/*^, and in this part of the 
lune will have a corresponding horizon- 
tal thrust. The curve e must also pass 
through /g and/g, but in this part of the 
lune will have a horizontal thrust cor- 
responding to it, differing from that in 
the part between f^ and f^^: indeed the 
horizontal thrust in the segment of a 
dome above any hoop depends exclusive- 
ly upon that segment and and is unaf- 
fected by the zone below the hoop. The 
tension sustained by the hoop is, how- 
ever, due to the radial force, which is 
the difference of the horizontal thrusts 
of the zones above and below the 
hoop. 

It is seen that the introduction of a 
second hoop will still further diminish 
the thickness of lune necessary to sus- 
tain the dome, unless indeed the thick- 
ness is required to sustain the meridian 
compression. 

Had a single hoop been introduced at 
f^ with none above that point, the dome 
above /*g should then be investigated, just 
as if the springing circle was situated at 
that point. The curve e must then start 
from /^, as it before did from /g, and be 
made to become tangent to the limit- 
ing curve at some point between /^ and 
the crown. 

By the method here employed for 
finding the tension of a hoop it is possi- 
ble to discuss at once the stresses in- 
duced in the important modern domes 
constructed with rings and ribs of metal 
and having the intermediate panels 
closed with glass. 

On introducing a large number of 
rings at small distances from each other, 
it will be seen that the discussion just 



given leads to the method previously 
given for the dome of metal. 

The dome of St. Paul's, London, is one 
which has excited much adverse criticism 
by reason of the novel means employed 
to overcome the difficulties inherent in so 
large a dome at so great a height above 
the foundations of the building. The 
exterior dome consists of a framework of 
oak sustained by conical dome of brick 
which forms the core. There is also a 
parabolic brick dome under the cone 
which forms no essential part of the sys- 
tem. Since the conical dome in general 
presents some peculiarities worthy of 
notice we will give an investigation of 
that form of structure as our concluding 
construction. 

CHAPTER XVH. 

CONICAL DOME OF METAL. 

In Fig. 19, let hd be the axis of the 
frustum of a metallic cone cut by a ver- 
tical plane in the meridian section a. 
The cone is supposed to have a uniform 
thickness too small to be regarded in 
comparison with its other dimensions. 
Suppose the frustum to be cut by a series 
of equi-distant horizontal planes as at g^^ 
g^, etc., into a series of frustra or rings : 
then the weight of each ring is propor- 
tional to its convex surface. The convex 
surface of any ring =-27trX slant height; 
when r is half the sum of the radii of the 
two bases, i.e., r is the mean radius. 
Consequently, the weights of these 
rings, or any given fraction of them in- 
cluded between two meridian planes, is 
proportional to their mean radii. Let us 
draw these mean radii d^a^, ^2^25 ^tc, be- 
tween the horizontals through g^^g^, etc., 
and use some convenient fraction, say ^, 
of these quantities of the type da as the 
weights. The line ii cuts off ^ of each 
of these : then lay off du^-=^d^i^ as the 
weight of the ring ag^, lay off u^i^-=^ 
^2^2? ^2^3^<^3^*35 etc., as the weights of 
the rings g^g^, g^g^, etc. 

Draw the line dt \\ aa, it corresponds 
to the curve st of Fig. 18; then the 
quantities of the type tu represent the 
horizontal radial thrust which the cone 
exerts upon the part below it, while the 
radial thrust borne by any ring is the 
difference between two successive quanti- 
ties of the type tu, i.e., the radial thrust 
in the ring g^g^ is represented by t^y^, 



60 



NEW COlSrSTRUCTIONS 



CONICAL DOIVIE 




that in g^g^ by t^y^^ etc. As previously 
shown in connection with the spherical 
dome, if the scale of weights be such 
that du^ represents a part of the cone 
between two meridian planes which make 
an angle of ^ = 180°-^;r=57''.3— , then 
will t^y^^ t^y^ etc., be the total hoop com- 
pression of the corresponding rings of 
the cone. It is to be noticed that this 
quantity does not change sign in the 
cone, and is always compression. 

The meridian compression is expressed, 
under the same circumstances by the 
quantities dt^^ dt^^ etc. 

Such a cone as this must be placed 
upon a cylindrical drum or other support 
which can exert a resistance in the direct- 
ion aa, but if this support is very 
slightly displaced by the horizontal radial 
thrust, a hoop tension will be induced at 
the base of the cone. As this displace- 
ment is very likely to occur it is far bet- 
ter to have the base of the cone sufficient- 
ly strong to withstand this tension, 
which is t^u^ when du^ is the weight of 
57°. 3 : then the supports will sustain a 
vertical force alone. 

This discussion applies equally well to 



a cone formed of a network of rings and 
inclined posts with intermediate panels 
of glass or other material. 

COl^^ICAL DOME OF MASOKRY. 

Let us assume that the uniform hori- 
zontal thickness of the dome to be 
treated, is one sixteenth of the internal 
diameter of the base, or one eighth of 
the internal radius, as shown in Fig. 19. 
The actual thickness is less than this, but 
since the horizontal thickness is a con- 
venient quantity, we shall call it the 
thickness unless otherwise specified. 

Pass equidistant horizontal planes as 
previously stated: then the volumes of 
these rings may be found by the pris- 
moidal formula. The volume 

in which h is the height of the ring, r^ 

and r/ are the radii external and internal 

of one base, r^ and r/ of the other, and r 

and r' of the middle section. Now 

r —r ':^r—r'=^r —rj^^t the thickness 
11 -1^ 

of the cone; and 



IN GEAPHICAL STATICS. 



61 



.*. Volume = 7tht{r '\-^')=^^7chtr 

when J (?• + /)=?' the mean radius of the 
middle section. From this it is seen 
that the weights vary in the same man- 
ner, and are represented by the same 
quantities as previously stated in case of 
a thin cone. Assume that the centers 
of gravity of any thin lunes cut from 
these rmgs by meridian planes making a 
small angle with each other, are at the 
middle points a^, a^, etc., this assumption 
is sufficiently exact for the part of the 
<3one near the base, which we are now 
specially to investigate. 

By means of the weights w^iu^=-ic^a^^ 
etc., at some assumed distance from the 
pole (/, describe the equilibrium polygon 
<?, starting from n at the inner third of 
the base. 

Now if the cone stands upon a drum 
which necessarily exerts a sufficient radi- 
al thrust to keep the meridian joints of 
the cone closed down to the base, then 
all the circumstances will be precisely as 
before explained in respect to the metallic 
dome : but if the drum exerts a less radi- 
cal thrust, the meridiab joints will open 
near the base, and the conditions of sta- 
bility of that part of the cone will need 
to be investigated, as was done in the 
spherical dome of masonry, by consider- 
ing the upper part of the dome as sus- 
tained by a series of stone arches. From 
/draw fc^ tangent to the curve c\ then 
must cfi^ be elongated to mfi^ and the 
other ordinates of c must be elongated 
in the same ratio in order that the equili- 
brium polygon may be tangent to the 
exterior limit fm; and, further, fm and 
fc^ are the ratio lines by which to effect 
the elongation. To find how much the 
thrust is diminished, draw through the 
intersection of /m with bd, a line parallel 
to fc^ intersecting the weight line at w, 
and then v the point where the horizontal 
through ID intersects /"m gives us the new 
position of the weight line, and its dis- 
tance from the pole d. This vertical in- 
tersects tt about midway between t^ and 
t,, thus showing that the meridian joints 
of the cone will be open from the base to 
about the point g^. It is unnecessary to 
draw the equilibrium polygon in its new 
position. 

We thus obtain the least horizontal 
thrust against which the dome can stand. 



The actual thrust which the drum exerts 
may have any value greater than this 
least thrust. 

It is seen that the effect of diminishing 
the thickness of the cone, is to carry the 
tangent point c^ and the point of no com- 
pression nearer to the base. In other 
words the *thin dome of masonry of given 
semi-vertical angle necessarily exerts a 
greater thrust in proportion to its weight 
than does a thick dome, though that 
proportion is unchanged if the joints 
are to remain closed all the way to the 
base. 

All of the circumstances respecting 
radial thrust above the point of no hoop 
compression, and respecting meridian 
thrust, are the same as in the metallic 
cone. 

Any additional loading above that of 
the weight of the cone itself, as for ex- 
ample, the weight of a lantern, or of an 
external dome, as in the case of St. Paul's, 
can be introduced and treated as an ad- 
ditional height or thickness of certain 
rings of the cone. The same method 
which has been here applied may be ap- 
plied to all such cases, if the weights be 
determined by some suitable process. 
For example, it may be shown by the 
help of the prismoidal formula, that the 
volume of the ring cut from a uniformly 
tapering cone by equidistant horizontal 
planes, varies as the product of the mean 
radius of the mid-section by the thick- 
ness at the mid-section. 

OTHER VAULTED STEUCTURES. 

Similar principles to those above devel- 
oped apply to domes with an elliptical or 
polygonal base, to domes whose meridian 
sections are ogee curves, to Skew Arches, 
to Groined Arches formed by the com- 
bination of cylindrical arches, as well 
as to Groined Arches which are dome- 
shaped. 

By the application of the principles 
developed it is easy to treat the cone or 
dome which sustains the pressure of earth 
or water. Indeed, it is not too much to 
say that the complete solution of the 
problem of the stability of vaulted struc- 
tures has now been set forth for the first 
time, and that the proper connection and 
relationship between similar structures, 
in metal and masonry, may now be 
clearly seen. In particular, the discus- 



62 



NEW COTSrSTEUCTIONS IN GEAPHICAL STATICS. 



sions have made manifest the applicabil- are all projections of any one of them, 
ity of a particular equilibrium polygon and the possil*lity of deriving from it in 
among the infinite number which are ' each of the structures treated, a complete 
due to a given set of weights, and which i and sufficiently exact solution. 



FIG. II 




62 



sions 
ity c 

am 01 
due ' 



FIG. Ill 




Fig.3 

ARCH RIB 

WITH FIXED ENDS 

and 

TTTNffiy. JOINT AT THE CROWN: 

GRAPHICAL METHOD. 

by 
HENRY T, EDDY, C.E. 



62 



sion 
ity 
amc 
due 



FIG. IV. 



Fig.4, 

TEMPERATURE STRAINS 

in the 
ST.LOUISARCH. 




62 



sion 
ity 
amc 
due 



FIG. V, 



Fig.5. 

TEMPERATURE STRAINS 




6J 



81 

it 
a] 
d 



FIG'S VI.VII.VIII.IX. 




Fig.9 

ARCH RIB 

WITH END AND CENTER JOINTS 
GRAPHICAL MF.THOD, 



FIG.X. 




A NEAV GENERAL METHOD 



IN 



GRAPHICAL STATICS 



A NEW GENERAL METHOD 



IN 



GRAPHICAL STATICS 



— *«"z:rjr ■■■ o 



All general processes used in the 
graphical computation of statical prob- 
lems consist, in their last analysis, in a 
systematized application of the proposi- 
tion known as the "parallelogram of 
forces," which states that if two forces 
be applied to a material point, and if 
they be represented in magnitude and 
direction by two determinate straight 
lines, then their resultant is represented 
in magnitude and direction by the 
diagonal of a parallelogram, two of 
whose sides are the just mentioned de- 
terminate lines. This is the basis of all 
grapKo-statical construction, but the 
methods by which it is systematized, and 
the auxiliary ideas incorporated in the 
processes, have so enlarged its possi- 
bilities of usefulness, that Graphical 
Statics may perhaps claim to be a science 
of itself; — the science of the geometrical 
treatment of force. 

In order to introduce to the public a 
new set of auxiliary ideas, which shall 
constitute a new method, of a character 
equally general with that now in use and 
known as the "equilibrium polygon 
method," it has seemed best to give, in 
the first place, a brief review of the prin- 
cipal ideas already employed by the cul- 
tivators of this science. 

RECIPROCAL FIGURES. 

When a framed structure, such as a 
roof or bridge truss, is subjected to the 
action of certain weights or forces, these 
applied forces form a system which is in 



] equilibrium. Now any system of forces 
in equilibrium may be represented in 
magnitude and direction by the sides of 
a closed polygon, a fact which follows 
at once from the doctrine of the parallelo- 
gram of forces. Such a polygon is called 
the polygon of the applied forces. 

Again, the forces which act at any 
joint of a frame are in equilibrium, and 
hence there is a closed polygon of the 
forces acting at each joint. The forces 
which meet at a joint of a frame are the 
longitudinal tensions or compressions of 
the pieces meeting at that joint, together 
with any of the applied forces whose 
point of application may be the joint in 
question. Draw a diagram of the frame 
and the applied forces all of which we 
will suppose lie in a single plane. Call 
this the "frame diagram:" it represents 
the position and direction of all the 
forces acting in and upon the frame. 
The frame diagram necessarily has at 
least three lines meeting at each joint. 
A piece which constitutes part of the 
frame does not necessarily have both 
its extremities attached at joints of the 
frame; one extremity may be firmly at- 
tached to any immovable object. The 
frame diagram is, therefore, not neces- 
sarily made up of closed figures. 

Now draw the closed polygon of the 
forces applied to the frame, and at each 
of the joints where forces are applied 
draw the closed polygon of the forces 
which meet at that joint, using so far as 
possible the lines already drawn as sides 



66 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



of the new polygons, and at the same 
time draw polygons for the forces acting 
at each of the remaining joints. If this 
process be effected with care as to the 
order of procedure, as well as to the 
order in which the forces follow each 
other in the polygon of the applied 
forces, then the resulting " diagram of 
forces," which is formed of the combi- 
nation of the polygon of the applied 
forces with the polygons for each joint, 
will contain in it a single line and no 
more parallel to each line of the frame 
diagram. In that case the force dia- 
gram is said to be a reciprocal figure to 
the frame diagram. If sufficient care is 
not exercised in the particulars men- 
tioned some of the lines in the force 
diagram will have to be repeated, and 
the figure drawn will not be the recipro- 
cal of the frame diagram, nevertheless 
it will give a correct construction of the 
quantities sought. 

If the frame diagram and the force 
diagram are both closed figures then 
they are mutually reciprocal. The 
properties of reciprocal figures were 
clearly set forth by Professor James 
Clerk Maxwell, in the Philosophical 



Magazine^ vol. 27, 1864; in which is 
stated, what is also evident from con- 
siderations already adduced above, that 
mutually "reciprocal figures are me- 
chanically reciprocal; that is, either may 
be taken as representing a system of 
points {i.e. joints) and the other as rep- 
resenting the magnitudes of the forces 
acting between them." 

The subject has also been treated by 
Professor B. Cremona in a memoir en- 
titled "Le figure reciproche nelle statica 
grafica." Milan, 1872. 

We shall now give examples of this 
method of computing the forces acting 
between the joints of a frame, together 
with certain extensions by which we are 
enabled to treat moving loads, etc. 
The method is correctly called " Clerk 
Maxwell's Method." The notation em- 
ployed, which is particularly suitable for 
the treatment of reciprocal diagrams, is 
due to R. H. Bow, C.E. ; and is used by 
him in his work entiled " Economics of 
Construction."^ London, 1873. In this 
work will be found a very large number 
of frame and force diagrams drawn by 
this method. 

Let the right hand part of Fig. 1 




represent a roof truss having an in- 
clination of 30° to the horizon, of 
which the lower chord is a polygon in- 
scribed in an arc of 60° of a circle. If 
the lower extremities of the truss abut 
against immovable walls a change of 
temperature causes an horizontal force 
between these lower joints, the effect of 
which upon the different pieces of the 
truss is to be constructed. No other 
weights or forces are now considered 
except those due to this horizontal force. 



FIg.l. 

ROOF TRUSS 

TEMPERATURE STRESSES 



This force is considered thus apart from 
all others because it is a force between 
two joints, and must enable us to obtain 
a pair of mutually reciprocal figures, 
such as weights and other applied forces 
seldom give. 

It is seen that the force between these 
joints might be suppobcd to be caused 
by a tie joining these points; and in 
general it may be stated that the dia- 
gram of forces due to any cambering or 
stress induced in a frame by "keying" 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



67 



pieces, is mutually reciprocal to the 
frame diagram. 

Let any piece of the frame be denoted 
by the letters in the spaces on each side 
of it; thus the pieces of the lower chord 
are qa^ qc, qe^ etc.; and those of the 
upper chord are rh, rd, etc., while ah, he, 
etc., are pieces of the bracing, and qr is 
the tie whose tension produces the stress 
under consideration. 

In the force diagram upon the left, let 
qr represent, on some assumed scale of 
tons to the inch, the tension in the piece 
qr ; and complete the triangle aqr with 
its sides parallel to the pieces which con- 
verge to the joint aqr; then must this 
triangle represent the forces which are 
in equilibrium at that joint. Kext, with 
ar as one side, complete the triangle ahr, 
by making its sides parallel to the pieces 
meeting at the joint of the same name: — 
its sides will represent the forces in 
equilibrium at that joint. In a similar 
manner we proceed from joint to joint, 
using the stresses already obtained in 
determining those at the successive 
joints. 

It is not possible to determine in 
general more than two unknown stresses 
in passing to a new joint, unless aided 
by some considerations of symmetry 
which may exist at such a joint as ghijq. 

Now from the left hand figure as a 
frame diagram, in w^iich stresses are 
induced by causing tension in the tie qr, 
we can construct the right hand figure 
as a force diagram, but it must be noticed 
in that case that rh, rh, rf, rd are sepa- 
rate and distinct pieces meeting at the 
joint r, although they all lie in the same 
right line, and that the same is true 
along the line oik on. 

One or two considerations of a general 
nature should be recalled in this con- 
nection. 

A polygon encloses the space q : in 
the reciprocal figure the lines parallel to 
its sides must all diverge from the point 
q: and if the upper chord had been a 
polygon, instead of being of uniform 
slope, the lines parallel to its sides would 
diverge from the point r. As it is, ra, 
rh, rd, rm etc., form the rays of such a 
pencil, in which several rays are super- 
posed one upon another. 

The determination of the question 
as to whether the stress in a given 
piece is tension or compression is 



effected by following the polygon for 
any joint completely around and noting 
whether the forces act toward or from 
the joint ; e.g, at the point fghrf, from 
following the diagrams of preceding 
joints in the manner stated, it will be 
found thatyj/ is under tension, and acts 
from the joint; consequently, gh which 
acts toward the joint is under compres- 
sion, as are also the two remaining pieces. 
Hence if the tension in the tie qr be re- 
placed by an equal compression in a part, 
tending to move the lower extremities 
of the roof from each other, the sign of 
every stress in the roof will be changed, 
but the numerical amount will remain 
unchanged, and no change will be made 
in the force diagram. 

ROOF TRUSS. 

As another example let us take a roof 
truss represented in Fig. 2, acted upon 
by the equal weights /e, ed, dd' , etc. 
Suppose that the effect of the wind 
against the right hand side of the truss 
is such as to cause a deviation of the 
force applied at the joint a'h'e'f of the 
amount indicated in the figure. Such a 
deviation may of course occur at several 
joints of a roof, but the treatment of 
the single joint at which the force of the 
wind is, in this case, principally concent 
trated, will sufliciently indicate the me- 
thod to be employed in more intricate 
examples. 

Suppose that this pressure of the wind 
is sustained by the left abutment. The 
manner in which it is really sustained 
depends upon the method by which the 
roof is fixed to the walls. 

This horizontal pressure of the wind is 
not directly opposed to the thrust of the 
left abutment, consequently a couple is 
brought into play by these forces, whose 
effect is to transfer a part of the weight 
from the right to the left abutment. To 
compute the amount of this effect, draw 
an horizontal line through this joint (or 
in case the wind acts at several joints the 
horizontal line has to be drawn through 
the center of action of the wind pressure) 
and prolong it until it intersects the 
vertical at the right abutment at 3. Let 
14 be equal to the pressure of the wind. 
Join 13 and prolong 13 until it intersects 
the vertical through 4 at 5, then is 45 
the amount by which the weight upon 
the left abutment is increased, and that 



68 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 




upon the left abutment decreased. For, 

let k.'T4='\2 . then^. 45 = 23. Now the 

couple due to the wind =23 • 14 but 

^~23.T4=T2. 23=^. 12". 45, .'. "23 . 

14=12 . 45. The right hand side of this 
last equation is the couple equivalent to 
the wind couple, having the arm 12 and 
a pair of equal and opposite forces repre- 
sented by 45. Let 45 be added to half 
the weight of the symmetrical loading 
upon the roof to obtain the vertical re- 
action of the left abutment, and sub- 
tracted from the same quantity for the 
vertical reaction of the right abutment. 

If any doubt occurs as to the manner 
in which the wind pressure is distributed 
between the abutments that distribution 
should be adopted which will cause the 
greatest stresses upon the pieces, or, as 
it may be stated in better terms, each 
piece should be proportioned to bear the 
greatest stress which any distribution of 
that pressure can cause. 

Let us suppose that a horizontal com- 
pression is exerted upon the truss due to 
temperature or other cause, and repre- 
sented by the width 26 of the rectangle 
at the right abutment, then the reaction 
at that point is the resultant 92 of this 
compression and the vertical reaction; 
while at the left abutment the total hori- 
zontal reaction 71 is the sum of this 
compression and the resistance called 
into action by the wind, giving 81 as the 
resultant reaction at the left abutment. 



Fig.2. 

TEMPERATURE, 

WIND AND WEIGHT STRESSES 



Now, using a scale of force twice that 
just employed, for the sake of greater 
convenience and accuracy, construct 
defyfe'd' the polygon of the applied 
forces; and proceed to construct as in 
Fig. 1 the polygons of forces for each of 
the joints. The accuracy of the conr 
struction will be tested by the closing 
of the figure at the completion of the 
process. 

The force diagram at the left is the 
reciprocal figure of the diagram of the 
frame and applied forces at the right, 
but the figure at the right is not the re- 
ciprocal of that at the left since it is not 
a closed figure with at least three lines 
meeting at each intersection. 

BRIDGE TRUSS. 

As a further example take the bridge 
truss shown in Fig. 3, which is repre- 
sented as of disproportionate depth in 
order to fit the diagram to the size of the 
page. The method employed is a simpli- 
fication of that given by Mr. Charles H. 
Tutton on page 385, vol. XVII of this 
Magazine. 

Let us suppose the dead load of the 
bridge itself to consist of a series of 
equal weights ?o, applied at the upper 
joints cCj, x'2, etc., of the bridge. Let 
each of these weights when laid off to 
scale be represented by the length of 
zy"'=zw, then the horizontal lines xx and 
y"'o include between them ordinates 
which represent these weights. 



A NEW GENERAL METHOD IN GKAPIIICAL STATICS. 



69 




^ 


6jj,&9&8 ^V^S ^5 *4 &:! 1^2 


>^ 


&2 


">. 


>4 


&, 


&6 


&7»| 




////// III/ 




/ 






/ 


/ 


/" 




VIIHI 1 1 




/ 






/ 


/ 




illllll 




/ 


1 




/ ^*7 


•^s 


c'J / 1 / / 




1 


1 


/ '"8 




IJ 1 1 1 1 




j 


j 




Vi* 


Ljjjl 




1 




Ci 


ij j 


j 


1 


Flg.3. 




III 


/ 


'«. 


BRIDGE TRUSS 


''3 


ill 


j 




1 


/ 




MAXIMUM STRESSES 




1 / 


/ 


\ 




''2 


cJ 






«. 


a, 






' 











Let the live load consist of one or 
more locomotives which stand at the 
joints x^ and x^, and a uniform train of 
cars which covers the remaining joints. 
Let the load at each joint due to the cars 
be represented by y"'y'-=w' ^ and the ex- 
cess above this of the load at each of the 
joints covered by the locomotives be 
represented hj y'y"=^v)\ .\ w + w' + io" 
the load at x^ and at 
=zy' is the load at x„ 



is 



z=G^c^=zy 

X.. and w-\-w' ^=^(iji^ 



= c^c^ 



and at each of the remaining joints. 

Draw y'o, y"o and 20, then is z^y" 
=^2?/* that part of the load at x^ 
which is sustained at the left abutment, 
as appears from the principle of the 
lever. Again ^22//'— Ti^2/" i'^ ^^^^ P^^'* 



of the load at x^ sustained by the same 
abutment, and ^^y^^^W^y' ''^^ a similar 
part of load at x^. Let the sum of these 
weights sustained by the left abutment 
be obtained; it is c^e upon the lower 
figure. Upon cfi lay off c^c^-^w-Vw* 
+ 1//, G^e^—w^w'^^^o\ c^e^=w-hw\ etc., 
equal to the loads applied at x^, x^, etc. 
We are now prepared to construct a dia- 
gram of forces which shall give the 
stresses in the various pieces uuder this 
assumed loading. Before constructing 
such a diagram, we wish to show that 
the assumed position of the load causes 
greater stresses in the chords of the 
bridge than any other possible position. 
The demonstration is quoted nearly ver- 



70 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



batim from Rankine's Applied Mechanics, 
and though not strictly applicable to the 
case in hand, since it refers to a uni- 
formly distributed load, it is substan- 
tially true for the loading supposed, 
when the excess of weight in the loco- 
motives is not greater than occurs in 
practice. 

"For a given intensity of load per 
unit of length, a uniform load over the 
whole span produces a greater moment 
of flexure at each cross section than any 
partial load." 

" Call the extremities of the span 1 
and 2, and any intermediate cross section 
3. Then for a uniform load, the moment 
of flexure at 3 is an upward moment, be- 
ing equal to the upward moment of the 
supporting force at either 1 or 2 rela- 
tively to 3, minus the downward moment 
of the uniform load between that end 
and 3. A partial load is produced by 
removing the uniform load from part of 
the span, situated either between 1 and 
3, between 2 and 3, or at both sides of 3. 
First, let the load be removed from any 
part of the span between 1 and 3. Then 
the downward moment, relatively to 3, 
of the load between 2 and 3 is unaltered, 
and the upward moment, relatively to 3, 
of the supporting force at 2 is diminished 
in consequence of the diminution of the 
force; therefore the moment of flexure 
is diminished. A similar demonstration 
applies to the case in which the load is 
removed from a part of the span be- 
tween 2 and 3; and the combined effect 
of those two operations takes place when 
the load is removed from portions of the 
span lying at both sides of 3; so that 
the removal of the load from any portion 
of the beam diminishes the moment of 
flexure at each point." 

The stress upon a chord multiplied by 
the height of the truss is equal to the 
moment of flexure; hence in a truss of 
uniform height the stresses upon the 
chords are proportional to the moments 
of flexure, and when one has its greatest 
value the other has also. 

The sides of the triangle c^eh^ repre- 
sents the forces in equilibrium at the 
joint c,e&, at the left abutment 1. The 
polygon c^Gf>^a^c^ represents the forces 
in equilibrium at the joint of the same 
name, ^.e., at the joint x^. The forces at 
the other joints are found in a similar 
manner. 



It is unnecessary to complete the 
flgure above e unless to check the 
process. The stresses obtained for the 
corresponding pieces in the right half of 
the truss would, upon completing the 
diagram, be found to be slightly less 
than those already determined because 
there are no locomotives at the right. 
The greatest stresses upon the pieces 
of the lower chord are eh^^ 6^,, etc., and 
on the upper chord are ^gCg, a,c„ etc. 

To determine the greatest stress upon 
the pieces of the bracing (posts and ties) 
it is necessary to find what distribution 
of loading causes the greatest shearing 
force at each joint, since the shearing 
forces are held in equilibrium by the 
bracing. We again quote nearly word 
for word from Rankine's Applied Me- 
chanics. 

"For a given intensity of load per 
unit of length, the greatest shearing 
force at any given cross-section in a 
span takes place when the longer of the 
two parts into which that section di- 
vides the span is loaded, and the shorter 
unloaded." 

" Call the extremities of the span, as 
before, 1 and 2, and the given cross- 
section 3; and let 13 be the longer part, 
and 23 the shorter part of the span. In 
the first place, let 13 be loaded and 23 
unloaded. Then the shearing force at 3 
is equal to the supporting force at 2, and 
consists of a tendency of 23 to slide up- 
wards relatively to 13. The load may be 
altered either by putting weight between 
2 and 3, or by removing weight between 

1 and 3. If any weight be put between 

2 and 3, a force equal to part of that 
weight is added to the supporting force 
at 2, and, therefore, to the shearing force 
at 3 ; but at the same time a force equal 
to the whole of that weight is taken away 
from that shearing force; therefore the 
shearing force at 3 is diminished by this 
alteration of the load. If weight be re- 
moved from the load between 1 and 2, 
the shearing force at 3 is diminished 
also, because of the diminution of the 
supporting force at 2. Therefore any 
alteration from that distribution of load 
in which the longer segment 13 is loaded, 
and the shorter segment 23 is unloaded, 
diminishes the shearing force at 3." 

The shearing force at any point is the 
resultant vertical force at that point, 
and can be computed by subtracting 



A NEW GENERAL METHOD IN aRAPHICAL STATICS. 



from the weight which rests upon either 
abutment the sum of all the weights be- 
tween that point and the abutment, i.e., 
by taking the algebraic sum of all the 
external forces acting upon the truss 
from either extremity to the point in 
question; the reaction of the abutment 
is, of course, one of these external 
' forces. 

The greatest stress upon the brace 
a^bj is that already found, while x^ is 
loaded with the live load. 

If the live load be moved to the right 
so that no live load rests upon x^, and 
the locomotives rest upon x^ and x^^ the 
pieces b^a^ and a^b^ will sustain their 
greatest stress. To find the shear at x^ 
in that case, we notice that the change 
in position of the live load has changed 
the reaction c^e of the left abutment by 
the following amounts : the reaction has 
been diminished by the quantity yl" yl^' 
=^ {w' -\-io'')^ since the load at x^ has 
been removed, and it has been increased 
by y ^' y ^" =^\^io" , since x^ is loaded more 
heavily than before, therefore the re- 
action of the abutment has on the whole 
been decreased by the total amount -^ 
(\bw' + 2w"). 

Now the shear at x^ is this reaction di- 
minished by the load lo at x^. In order 
to construct it, draw 2/2/14" parallel to 
2/'o, then yy'z=i^%o" . .*. Shear at x^ 

=Lec^ — w — yig- [ibw' -^210") = ec^ — x^y^. 
Lay off c^c^^=^x^y^^ then the shear at 
x^ = ec/ = the greatest stress in the 
brace b^a^; and b^c^=. the greatest stress 
in alj^. 

Again, to find the greatest shear at x^ 
when the live load has moved one panel 
further to the right, we have the equa- 
tion: Shear at x^^^ec^ —w—\^ {w' + io") 

—x^^ Lay off c^c^'=x^y^, then the 
shear at x^=ec^', which is the greatest 
stress in the piece b^a^^ while b^'c^' is the 
greatest stress in aj)^. 

In similar manner lay off, c^'c^^^x^y^^ 
G^c^'^x^y^^ etc., until the whole of the 
original reaction ec^ of the abutment is 
exhausted, then are eCj, ec/, ec^', ec/, etc., 
the successive shearing stresses at the 
end of the load, i.e. the greatest shearing 
stresses, and consequently these stresses 
are the greatest stresses on the succes- 
sive vertical members of the bracing, 
while Cj6j, c/6/, c^'b^'^ etc., are the great- 



est stresses on the successive inclined 
members of the bracing. 

Had the greater load, such as the loco- 
motives, extended over a larger number 
of panels, the line y^y^y^ would have cut 
off a larger fraction of y'y" . Suppose, 
for instance, that the locomotives had 
covered the joints x^x^ inclusive, then 
the line y^y^ would have passed through 
^/g", and been parallel to its present posi- 
tion. In that case the ordinates x^y^, 
x^y^ would have been successively sub- 
tracted from the reaction of the abut- 
ment due to a live load covering every 
joint, in order to obtain the shearing 
forces, just as at present, until we arrive 
at iCg, after which it would be necessary 
to subtract the ordinates x^y/, ^g^/g", etc. 
The counter braces are drawn with 
broken lines. Two counters are necessary 
on each side of the middle under the 
kind of loading which we have supposed. 
It is convenient, and avoids confusion in 
lettering the diagram to let ci^b^, for in- 
stance, denote the principal or counter 
indifferently, as both are not subject to 
stress at the same time. 

The devices here used can be applied 
to a variety of cases in which the loading 
is not distributed in so simple a manner 
as in this case. 

IlSr GENERAL. 

This method permits the determina- 
tion of the stresses in any frame when 
we know the relative position of its 
pieces and the applied forces, provided 
the disposition of the pieces is such as to 
admit of a determination of the stresses. 

The determination of what the applied 
forces are in case of a continuous girder 
or arch i^ a matter of some complexity, 
depending upon the elasticity of the ma- 
terials employed, and the method in its 
present form affords little assistance in 
finding them. 

Some authors have applied the method 
to find the stresses induced in the various 
pieces of a frame by a single force first 
applied at one joint, and then at another, 
and so on, and, finally, to find the 
stresses induced by the action of several 
simultaneous forces, by taking the alge- 
braic sum of their separate effects. This 
is theoretically correct but laborious in 
practice in ordinary cases. Usually, some 
supposition respecting the applied forces 
can be made from which the results of 



73 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



all the other suppositions which must be 
made, can be derived with small labor. 
The bridge trass treated was a remarka- 
ble case in point. 

WHEEL WITH TENSION-ROD SPOKES. 

A very interesting example is found 
in the wheel represented in Fig. 4, in 
which the spokes are tension rods, and 



the rim is under compression. Let the 
greatest weight which the wheel ever sus- 
tains be applied at the hub of the wheel 
on the left, and let this weight be rep- 
resented by the force aa' on the right, 
which is also equal to the reaction of 
the point of support upon which the 
wheel stands; hence aa' represents the 
force acting between two joints of this 




frame. The same effect-would be caused 
upon the other members of the frame by 
"keying" the rod aa' sufficiently to 
cause this force to act between the hub 
and the lowest joint. 

It should be noticed in passing, that 
the weights of the parts of the wheel it- 
self are not here considered; their effect 
will be considered in Fig. 5. Also, the 
construction is based upon the supposi- 
tion that there is a flexible joint at the 
extremity of each spoke. This is not an 
incorrect supposition when the flexibility 
of the rim is considerable compared with 
the extensibility of the spokes, a condi- 
tion which is fulfilled in practice. 

A similar statement holds in the case 
of the roof truss with continuous rafters, 
or a bridge truss with a continuous upper 
chord. The flexibility of the rafters or 
the upper chord is sufficiently great in 
comparison with the extensibility of the 
bracing, to render the stresses practically 
the same as if pin joints existed at the 
extremities of the braces. 

Furthermore, the extremities of the 
spokes are supposed to be joined by 
straight pieces, since the forces be- 



tween the joints of the rim act in those 
directions. Such forces will cause small 
bending moments in the arcs of the rim 
joining the extremities of the spokes. 
Each arc of the rim is an arch subjected 
to a force along its chord or span, and it 
can be treated by the method applicable 
to arches. This discussion is unimport- 
ant in the present case and will be 
omitted. 

Upon completing the force polygon in 
the manner previously described, it is 
found that the stress on every spoke is 
the same in amount, and is represented 
by a side of the regular polygon abcd^ 
etc. upon the left, while the compression 
of the pieces of the rim are represented 
by the radii oa oh, etc. 

As previously explained these dia- 
grams are mutually reciprocal, and it 
happens in this case that they are also 
similar figures. 

We then conclude that in designing 
such a wheel each spoke ought to be 
proportioned to sustain the total load, 
and that the maker should key the 
spokes until each spoke sustains a stress 
at least equal to that load. Then in no 



A NEW GEjSLERAL METHOD IN GKAPIIICAL STATICS. 



7a 




position of the wheel can any spoke be- 
come loose. The load here spoken of 
includes, of course, the effect of the 
most severe blow to which the wheel 
may be subjected while in motion. 

WATER WHEEL WITH TENSION-EOD SPOKES. 

The effect of a load distributed uni- 
formly around the circumference of such 
a wheel as that just treated is repre- 
sented in Fig. 5. Should it be desirable 
to compute the effect of both sets of 
forces upon the same wheel, it will be 
sufficient to take the sum of the separate 
effects upon each piece for the total 
effect upon that piece, though it is 
perfectly possible to construct both at 
once. 

We shall suppose a uniform distribu- 
tion of the loading along the circumfer- 
ence in the case of the Water Wheel, 
because in wheels of this kind such is 
practically the case so far as the spokes 
are concerned, since the power is trans- 
mitted, not through them to the axis, 
but, instead, to a cog wheel situated near 
the centej* of gravity of the " water arc." 
This arrangement so diminishes the 
necessary weight of the wheel, and the 
consequent friction of the gudgeons, as 
to render its adoption very desirable. 



aa cannot 
' is suited 



The discussion of the stresses appears 
however, to have been heretofore^erro- 
neously made.* 

Let the weight pp\ at the highest 
joint of the wheel, be sustained by the 
rim alone, since the spoke 
assist in sustaining pp\ as aa 
to resist tension only. Conceive, for the 
moment, that two equal and opposite 
horizontal forces are introduced at the 
highest joint such as the two parts of 
the rim exert against each other, then 
^pp' ^^pqz=ip' q' being sustained by each 
of the pieces a/9, a'p' respectively we 
have apq and a'p'q' as the triangles 
which together represent the forces at 
the highest joint. The force aa' on the 
right is the upward force at the axis, 
equal and opposed to the resultant of 
the total load upon the wheel, and the 
apparent peculiarity of the diagram is 
due to this; — the direction of the reaction 
or sustaining force of the axis passes 
through the highest joint of the wheel 
and yet it is not a force acting between 
those joints and could not be replaced 
by keying the tie connecting those joints.- 
In other particulars the force diagram is 



* " A Manual of the Steam Engine, etc.," by W, 
Rankine. Page 18-2, 7th Ed. 



J. M 



74 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



constructed as previously described and 
is sufficiently explained by the lettering. 
Should the spoke aa' have an initial ten- 
sion greater than pp' ^ then there is a 
residual tension due to the difference of 
those quantities whose effect must be 
found as in Fig. 4. 

Should the wheel revolve with so great 
a velocity that the centrifugal force 
must be considered, its effect will be to 
increase the tension on each of the spokes 
by the same amount, — the amount due 
to the deviating force of the mass sup- 
posed to be concentrated at the extremity 
of each spoke. The compression of the 
rim may be decreased by the centrifugal 
force, but as this is a temporarj'^ relief, 
occurring only during the motion, it does 
not diminish the maximum compression 
to which the rim will be subjected. 

We conclude then, that every spoke 
must be proportioned to endure a ten- 
sion as great as hh' from the loading 
alone; and that if other forces, due to 
centrifugal force or to keying, are to act 
they must be provided for in addition. 
Furthermore, we see that the rim must 
be proportioned to bear a compression 
as great as /^^, due to the loading alone, 
and that the centrifugal force will not 
increase this, but any keying of the 
spokes beyond that sufficient to produce 
an initial tension on each spoke as great 
as pp' must be provided for in addi- 
tion. 

The diagram could have been con- 
structed with the same facility in case 
the applied weights had been supposed 
unequal. 

It can be readily shown that the dif- 
ferential equation of the curve circum- 
scribing 'the polygon abcd^ etc. of Fig. 5 
is 

dx —^ldx\ 



dy 



which equation is not readily integrable. 
When, however, the number of spokes is 
indefinitely increased, it appears from 
simple geometrical considerations that 
this curve becomes a cycloid having its 
cusps at q and q' . 

ASSUMED FRAMING. 

Thus far, we have treated the effect 
of known external forces upon a given 
form of framing, and it is evident from 
the previous discussions and the illustra- 



tive examples that any such problem, 
which is of a determinate nature, can be 
readily solved by this method. But in 
case the problem under discussion has 
reference to the relations of forces among 
themselves, it is necessary to assume 
that the forces are applied to a frame or 
other body, in order to obtain the re- 
quired relationship. Certain general 
forms of assumed framing have proper- 
ties which are of material assistance in 
treating such problems, and this is true 
to such an extent that even though the 
form of framing to which the forces are 
applied is given, it is still advantageous 
to assume, for the time being, one of the 
forms having properties not found in 
ordinary framing. The special framing 
which has been heretofore assumed for 
such purposes is the Equilibrium Polygon, 
whose various properties will be treated 
in order. We now propose another form 
of framing, which we have ventured to 
call the Frame Pencil, with equally 
advantageous properties which will also 
be treated in due order. 

It may be mentioned here, that the 
particular case of parallel forces is that 
most frequently met with in practice. In 
case of parallel forces the properties of 
the equilibrium polygon and frame pen- 
cil are more numerous and important 
than those belonging to the general case 
alone. We shall first treat the general 
case, and afterwards derive the additional 
properties belonging to parallel forces. 

THE EQUILIBRIUM POLYGON FOR ANY 
FORCES IN ONE PLANE. 

Let a5, 5c, cd^ de Fig. 6 be the dia- 
gram of any forces lying in the plane of 
the paper, and ahcde their force polygon, 
then, as previously shown, ae the ch>sing 
side of the polygon of the applied forces 
represents the resultant of the given 
forces in amount and direction. Assume 
any point /> as a pole, and draw the 
force "^ewoW p— abode. The object in view 
in so doing, is to use this force pencil 
and polygon of the applied forces 
together in order to determine a figure 
of which it is the reciprocal. 

From any convenient point as 2 draw 
the side ap parallel to the ray ap until 
it intersects the line of action of the force 
ab^ and from that intersection draw the 
side bp parallel to the ray bp^ etc., etc.; 
then the polygon jt) will have its sides 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



75 



EQUILIBRIUM POLYGON. 




RECIPROCAL FIGURES. 



f Force Diagram, abcde, 

DwecUcn and \ Equilibrium Polygon, ap, bp, cp, dp, ep, 
■I Equilibrium Polygon, ap' , bp' , cp' , dp' , ep' , 
PoBiiion. Closing Line, 23 || pq. 

Resultant Force, ae. 



Force Polygon. 
Foice Pencil. 
Force Pencil. 
Closing Bay. 
Resultant Force. 



Direction and 
Magnitude. 



parallel respectively to the rays of the 
pencil J9. 

The polygon p and the given forces 
a5, &c, etc, then form a force and frame 
diagram to which the pencil j9— a JccZe is 
reciprocal, and of which it is the force 
diagram. It is seen that no internal 
bracing is needed in the polygon jo, and 
hence it is called an equilibrium (frame) 
polygon: it is the form which a funicular 
polygon, catenary, or equilibrated arch, 
would assume if occupying this position 
and acted upon by the given forces. 

As represented in Fig. 6 the sides of 
the polygon jt? are all in compression so 
that p represents an ideal arch. If the 
line 23 be drawn cutting the sides ap, ep 
so that it be considered to be the span of 
the arch having the points of support 2 
ancl 3, then this arch exerts a thrust in 
the direction 23 which may be borne 
either by a tie 23 or by fixed abutments 
2 and 3 : the force in either case is the 



same and is represented by pq \\ 23. It 
is usual to call 23 a closing line of the 
polygon p. The point q divides the 
resultant ae into two parts such that 
qapq and epqe are triangles whose sides 
represent forces in equilibrium, i.e., the 
forces at the points 2 and 3; hence, qa 
and eq are the parts of the total resultant 
which would be applied at 2 and 3 
respectively. 

This method is frequently employed 
to find the forces acting at the abutments 
of a bridge or roof truss such as that in 
Fig. 2. But it appears that it has often 
been erroneously employed. It must be 
first ascertained whether the reaction at 
the abutments is really in the direction 
ae for the forces considered. It may 
often happen far otherwise. If the 
surfaces upon which the truss rests with- 
out friction are perpendicular to ae, then 
this assumption is probably correct; as, 
for instance, when one end is mounted 



76 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



on rollers devoid of friction, running 
on a plate perpendicular to ae. But in 
cases of wind pressure against a roof 
truss the assumption is believed to be in 
ordinary cases quite incorrect. Indeed, 
the friction of the rollers at end of a 
bridge has been thought to cause a 
material deviation from the determina- 
tion founded on this assumption. It is 
to be noticed that any point whatever on 
pq {or pq prolonged) might be joined to 
a and e for the purpose of finding the re- 
actions of the abutments. Call such a 
point X (not drawn), then ax and ex might 
be taken as two forces which are exerted 
at two and 3 by the given system. It ap- 
pears necessary to call attention to this 
point, as the fallacious determination of 
the reactions is involved in a recently 
published article upon this subject.* We 
shall return to the subject again while 
treating parallel forces and shall extend 
the method given in connection with 
Fig. 2 to certain definite assumptions, 
such as will determine the maximum 
stresses which the forces can produce. 

Prolong the two sides ap and ep of the 
polygon jo until they meet. It is evident 
that if a force equal to the resultant ae be 
applied at this intersection of ap and ep 
prolonged, then the triangles apq and epq 
will represent the stresses produced at 2 
and 3 by the resultant. But as these aro 
the stresses actually produced by the 
forces, and as the resultant should cause 
the same effects at 2 and 3 as the forces, 
it follows that the intersection of ap and 
ep must be a point of the resultant ae ; 
and if, through this intersection, a line 
be drawn parallel to the resultant ae, it 
will be a diagram of the resultant, 
showing it in its true position and 
direction. 

This is in reality a geometric relation- 
ship and can be proved from geometric 
considerations alone. It is sufficient for 
our purposes, however, to have estab- 
lished its truth from the above mentioned 
static considerations which may be re- 
garded as mechanical proof of the 
geometric proposition. 

The pole p was taken at random : let 
any other point jo' be taken as a pole. 
To avoid multiplying lines p' has been 



• See paper No. 71 of the Civil Engfineers' Club of the 
Northwest. Applications of the Equilibrium Polygon 
to determine the Reactions at the Supports of Roof 
Trusses. By James R, Willett, Architect. Chicago. 



taken upon pq. Now draw the force 
pencil p' — abode and the co:f responding 
equilibrium polygon for the same forces 
ah, he, etc. This equilibrium polygon 
has all its pieces in tension except p'c. 
It is to be noticed that the forces are 
employed in the same order as in the 
previous construction, because that is the 
order in the polygon of the applied 
forces : but the order of the forces in 
the polygon of the applied forces is, at 
the commencement, a matter of indiffer- 
'^ence, for the construction did not depend 
upon any particular succession of the 
forces. 

As previously shown, the intersection 
of ap' with ep' i'^ a point of the result- 
ant, and the line joining this intersection 
with the corresponding intersection 
above is parallel to ae. 

Again, prolong the corresponding sides 
of the two equilibrium polygons until 
they intersect at 1234, these points fall 
upon one line parallel to pp^. For, sup- 
pose the forces which are applied to the 
lower polygon p' to be reversed in direc- 
tion, then the system applied to the poly- 
gons jo and p' must together be in equili- 
brium; and the only bracing needed is a 
piece 23 \\pp', since the upper forces pro- 
duce a tension pq along it, and the lower 
forces a tension qp' , while the parts aq 
and qe of the resultant which are applied 
at 2 and 3 are in equilibrium. The same 
result can be shown to hold for each of the 
forces separately; e.g. the opposite forces 
ah may be considered as if applied at 
opposite joints of a quadrilateral whose 
remaining joints are 1 and 2 : the force 
polygon corresponding to this quadrilat- 
eral is aphp' , hence 12 \\pp' . Hence 
1234 is a straight line. The intersection 
of pe and p'c does not fall within the 
limits of the figure. 

It is to be noticed that the proposi- 
tion just proved respecting the col- 
linearity of the intersections of the 
corresponding sides of these equili- 
brium polygons is one of a geometric 
nature and is susceptible of a purely 
geometric proof. 

THE FRAME PENCIL FOR ANY FORCES IN 
ONE PLANE. 

Let ah, he, cd, de in Fig. 7 represent a 
system of forces, of which ahcde is the 
force polygon. Choose any single point 
upon the line of action of each of these 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



77 



FRAME PENCILS. 




Direction and 
Position. 



RECIPROCAL FIQURES. 

f Force Diagram, ah c d e, Force Polygon. 

j F'ame Pencil, a' b' c' d' e' , Equilibrating Polygon. 

\ Frame Pencil, a" b" c" d" e", Equilibrating Polygon. 

Frame Polygon, bb' , cc' , dd' , ee\ Force Lines. 

Resultant Force, a e, Resultant Force. 

^ Resultant Ray, a' e , Resultant Side. 



Direction and 
Magnitude. 



J 



forces, and join these points to any as- 
sumed vertex v' by the rays of the frame 
pencil a'h'o^d'e' . Also join the success- 
ive points chosen by the lines hh' ^ cc\ dd' 
which form sides of what we shall call 
the frame polygon. Now consider the 
given forces to be borne by the frame 
pencil and frame polygon as a system of 
bracing, which system exerts a force at 
the vertex v' in some direction not yet 
known, and also exerts a force along 
some assumed piece ee', which may be 
regarded as forming a part of the frame 
polygon. The stresses upon the rays of 
the frame pencil will be represented by 
the sides of ah'c'd'e' which we shall call 
the equilibrating (force) polygon; while 
the stresses in the frame polygon are 
given by the force lines hh\ cc', etc. If a 
resultant ray a'e' be drawn from v' par- 
allel to the resultant side ae' of the 
equilibrating polygon it will intersect ee' 
at a point of the resultant of the system 



of forces; for that is a point at which if 
the resultant be applied it will cause the 
same stresses along the pieces ct'e' and ee' 
which support it as do the forces them- 
selves. 

If the point e' in the force polygon be 
moved along e'd', the locus of the inter- 
section of the corresponding positions of 
the resultant ray a'e' and the last side ee' 
will be the resultant ae. It would have 
been unnecessary to commence the equi- 
librating polygon at a had the direction 
of aa' been known. Having obtained 
the direction of aa' as shown at 8, the 
equilibrating polygon could be drawn 
by commencing at any point of aa, || 
aa'. 

In cases like that in the Fig., where 
there is no reason for choosing the points 
which determine the sides of the frame 
polygon otherwise, it is simpler to make 
the frame polygon a straight line, which 
may in that case be called the frame 



78 



A NEW GENERA.L METHOD IN GRAPHICAL STATICS. 



line. Then the force lines are parallel 
to each other and to aa' also. This is a 
practical simplification of the general 
case of much convenience. 

It should be noticed here that the 
equilibrium polygon, as well as the 
straight line, is one case of the frame 
polygon. The interesting geometric re- 
lationships to be found by constructing 
the frame and equilibrium polygons as 
coincident must be here omitted. 

Suppose that it is desired to find the 
point q which divides the resultant into 
two parts, which would be applied in 
the direction of the resultant at two 
such points as 8 and 9: draw a6 || v'% 
and 6'6 || v'9 and then through 6 draw 
qq' 11 89. This may be regarded as the 
same geometric proposition, which was 
proved when it was shown that the locus 
of the intersection of the two outside 
lines of the equilibrium polygons (recip- 
rocal to a given force pencil) is the re- 
sultant, and is parallel to the closing side 
of the polygon of the applied forces. 
The proposition now is, that the locus of 
the intersection of the two outside lines 
of the equilibrating polygon (reciprocal 
to a given frame pencil) is the resolving 
line, and is parallel to the abutment 
line: for these two statements are geo- 
metrically equivalent. 

Assume a different vertex -u", and 
draw the frame pencil and its correspond- 
ing equilibrating polygon a"b"c"d"e. If 
a, 6 and e 5 be drawn parallel to v" 8 
and v" 9 respectively their intersection 
is upon qq' as before proven. 

Again, the corresponding sides of these 
two equilibrating polygons intersect at 
12 3 4 upon a line parallel to v'v", for 
this is tjhe same geometric proposition 
respecting two vertices and their equili- 
brating polygons which was previously 
proved respecting two poles and their 
equilibrium polygons. 

It would be interesting to trace the 
geometric relations involved in different 
but related frame polygons, as for exam- 
ple, those whose corresponding sides in- 
tersect upon the same straight line, but 
as our present object is to set forth the 
essentials of the method, a consideration 
of these matters is omitted. Enough 
has been proven, however, to show that 
we have in the frame pencil an inde- 
pendent method equally general and 



fruitful with that of the equilibrium 
polygon. 

EQUILIBEIUM POLYGON FOR PARALLEL 

FORCES. 

Let the system of parallel forces in 
one plane be four in number as repre- 
sented in Fig. 8, viz : lo^w^, ^^'^s* ^^^-j 
acting in the verticals 2 3 4 5 of the 
force diagram on the left. Let the 
points of support be in the verticals 1 
and 6. 

The force polygon at the right re- 
duces, in case of vertical forces, to a ver- 
tical line wio. Assume any arbitrary 
point j!9 as pole of this force polygon, (or 
weight line, as it is often designated) 
and, parallel to the rays of the force 
pencil at p, draw the sides of the equili- 
brium polygon ee, in the manner pre- 
viously described. Draw the closing 
line hk of this polygon ee, and parallel 
to it draw the closing ray pq\ then, as 
previously shown, pq divides the result- 
ant w^w^ at q into two parts which are 
the reactions of the supports. The 
position of the resultant is in the vertical 
171771 which passes through the inter- 
section of the first and last sides of the 
polygon ee, as was also previously 
shown. 

Designate the horizontal distance from 
p to the weight line by the letter H. It 
happens in Fig. 8 that jt?t^j=H, but in 
any case the pole distance H is the hori- 
zontal component of the force pq acting 
along the closing line. 

Now by similarity of triangles 

the moment of flexure, or bending mo- 
ment at the vertical 2, which would be 
caused in a simple straight beam or gir- 
der under the action of the four given 
forces and resting upon supports in the 
verticals 1 and 6. 

Again, from similarity of triangles, 

KK (=62/3) =^3/3 : ''S:iw^w^ 

.'.S(kJ-ej:) = H.k,e, 

_ =qw^,hJi^—w^w^,hJi=M, 

the moment of flexure of the simple gir- 
der at the vertical 3. 

Similarly it can be shown in general 
that 

JEr.he=M, 



A NEW GENERAL METHOD IN GKAPHICAL STATICS. 



79 



EQUILIBRIUM POLYGON. 




i.e. that the moment of flexure at any 
vertical whatever (be it one of the 
verticals 2 3 4, etc., or not) is equal 
to the product of the assumed pole 
distance S multiplied by the vertical 
ordinate he included between the equili- 
brium polygon ee and the closing line 
kk at that vertical. 

From this it is evident that the 
equilibrium polygon is a moment curve, 
i.e. its vertical ordinate at any point 
of the span is proportional to the 
bending moment at that point of 
a girder sustaining the given weights 
and supported by simply resting without 
constraint upon piers at its extremities. 

From this demonstration it appears 
that H.e^f^-=xo^w^JiJi^ is the moment of 
the force w^io^ with respect to the verti- 
cal 3; and similarly H.v%^r)%^-=^w^w^.e^r}%^ 
is the moment of the same force with 
respect to the vertical through the cen- 
ter of gravity. Also, JE[.y^y^-=w^^.hJi^ 
is the moment of the same force with 
respect to the vertical 6. 

Similarly m^r)i^ is proportional to the 
moment of all forces at the right, and 
m^m^ to all the forces left of the center 
of gravity, \ivXm^m^-\-m^m^-=^^.^ as should 
be the case at the center of gravity, 
about which the moment vanishes. 
From these considerations it appears 
that the segments 'mm of the resultant 



I are proportional to the bending moments 

of a girder supporting the given weights 

I and resting without constraint upon a 

! single support at their center of gravity. 

i Let us move the pole to a new position 

\p' having the same pole distance H %% p^ 

and in such a position that the new clos- 

j ing line will be horizontal, i.e. p'q must 

be horizontal. 

One object in doing this is to furnish 
a sufficient test of the correctness of the 
drawing in a manner which will be im- 
mediately explained; and another is to 
transfer the moment curve to a new 
position CO such that its ordinates may 
be measured from an assumed horizontal 
position hh of the girder to which the 
forces are applied, so that the girder 
itself forms the closing line. 

The polygon cc must have its ordinates 
he equal to the corresponding ordinates 
^■e, for 

M^H.ke^H.hc 

Also the segments of the line mm are 
equal to the corresponding segments of 
the line nn for similar reasons. 

Again, as has been previously shown, 
the corresponding sides (and diagonals 
as well) of the polygons ee and ce inter- 
sect upon the line yy \\pp'. 

These equalities and intersections fur- 
nish a complete test of the correctness of 
the entire construction. 



80 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



FRAME PENCIL. 
Fig. 9 




FEAME PENCIL FOR PARALLEL FORCES. 

Let the same four parallel forces in 
one plane which were treated in Fig. 8 
be also treated in Fig. 9, and let them 
be applied at 2, 3, 4, 5 to a horizontal 
girder resting upon supports at 1 and 6. 

Use 16 as the frame line and choose 
any vertex v at pleasure from which to 
draw the frame pencil dd. Draw the 
force lines wd parallel to the horizontal 
frame line 16, and then draw the equili- 
brating polygon dd with its sides paral- 
lel to the rays of the frame pencil dd. 

As has been previously shown, if a re- 
sultant ray vo of the frame pencil dd be 
drawn from v, as represented in Fig. 9, 
parallel to the closing side uu of the 
equilibrating polygon, this ray intersects 
16 at the point o where the resultant of 
the four given forces cuts 16. 

Furthermore, the lines w^r^ and d^r^ 
parallel to the abutment rays -yl and vQ 
of the frame pencil intersect on rr the 
resolving line, which determines the 
point of division q of the reactions of 
the two supports, as was before shown. 

Let the vertical distance between the 
vertex and the frame line be denoted by 
V, 

In Fig. 9 it happens that v6= V. 
If the frame polygon is not straight, or 
being straight is inclined to the horizon, 



F'has different values at the different 
joints of the frame polygon: in every 
case V is the vertical distance of the 
joint under consideration above or below 
the vertex. It will be found in the se- 
quel that this possible variation of V 
may in certain constructions be of con- 
siderable use. 

By similarity of triangles we have 

12 : v6 : : r^r^ : w^q 



■■M. 



2> 



.*. Y.r^r^=w^q.l2 

the bending moment of the girder at the 
point 2. 

Draw a line through w^ parallel to v3; 
this line by chance coincides so nearly 
with w^s^ that we will consider that it is 
the line required, though it was drawn 
for another purpose. Again, by simi- 
larity of triangles 

13 : v6 : : r^s^ : w^q 
23 : V6 : : d^g{=r^s^) : w^w^ 

= tv^q.l3 — w^w^.23=M^ 
the bending moment at 3. 

Similarly it may be shown that 

i.e. that the moment of flexure at any 
point of application of a force to the 
girder is the product of the assumed 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



81 



vertical distance V multiplied by the 
corresponding segment rr of the resolv- 
ing line. 

The moment of flexure at any point 
of the girder may be found by drawing 
a line tangent to the equilibrating poly- 
gon (or curve) parallel to a ray of the 
frame pencil at that point, the intercept 
r,r of this tanajent is such that Fir,r is 

1 O 1 

the moment required. 

Also by similarity of triangles 

02 : vQ : : u^d^ : w^w^ 

.', VM^d^=w^w^.o2 

o2(=o3-f32) : vQ : : u^l : lo^w^ 

32 : v6 : : dj : w^w^ 

••• v{uj-d,i)=rMA 

i.e. the horizontal abscissas ud between 
the equilibrating polygon dd and its 
closing side uu multiplied by the verti- 
cal distance V are the algebraic sum of 
the moments of the forces about their 
center of gravity. The moment of any 
single force about the center of gravity 
being the difference between two success- 
ive algebraic sums may be found thus: 
draw dj || uu, then is V^.d/ the moment 
of w^w^ about the center of gravity, as 
may be also proved by similarity of tri 
angles. 

Again by proportions derived from 
similar triangles, precisely like those 
already employed, it appears that 

Y.w^d^=w^w^.2Q 

is the moment of the force w^w^ about 
the point 6. And similarly it may be 
shown that 

V.w^d^ =w^w^.2Q + «^^^3. 3 6 

is the moment of w^tu^ and w^w^ about 6. 
Furthermore, as this point 6 was not 
specially related to the points of applica- 
tion 1 2 3 4, we have thus proved the 
following property of the equilibrating 
polygon: if a pseudo resultant ray of 
the frame pencil be drawn to any point 
of the frame line, then the horizontal 
abscissas between the equilibrating poly- 
gon and a side of it parallel to that ray, 
(which may be called a pseudo closing 
side), are proportional to the sum total 
of the moments about that point of those 
forces which are found between that 
abscissa and the end of the weight line 
from which this pseudo side was drawn. 
The difference between two successive 



sum totals being the moment of a single 
force, a parallel to the pseudo side en- 
ables us to obtain at once the moment of 
any force about the point, e.g. draw d^i^ 
II wio .*. V.dJ' is the moment of w^io^ 
about 6. 

Now move the vertex to a new posi- 
tion v' in the same vertical with o : this 
will cause the closing side of the equili- 
brating polygon (parallel to v'o) to coin- 
cide with the weight line. The new 
equilibrating polygon bb has its sides 
parallel to the rays of the frame pencil 
whose vertex is at v\ If V^ is un- 
changed the abscissas and segments of* 
the resolving line are unchanged, and vv' 
is horizontal. Also xx \\ vv' contains 
the intersections of corresponding sides 
and diagonals of the equilibrating poly- 
gon. These statements are geometri- 
cally equivalent to those made and 
proved in connection with the equili- 
brium polygon and force pencil. 

In Figs. 8 and 9 we have taken j5^= F, 
hence the following equations will be 
found to hold, 



^>,e=r.r^^, 7c^e=r^r^, Jc^e=r^r^, etc. 



m^m^—u^d^, 7n^m^=u^d^, m^m^=u^d^, etc. 

y,y,='^^<^,, 2/i2/3=^3^3J yi2/4^^4^4; etc. 
m^m=dj,, etc., yjc=dj.', etc. 

By the use of etc. we refer to the more 
general case of many forces. From 
these equations the nature of the rela- 
tionship existing between the force and 
frame pencils and their equilibrium and 
equilibrating polygons becomes clear. 
Let us state it in words. 

The height of the vertex (a vertical 
distance), and the pole distance (a hori- 
zontal force) stand as the type of the 
reciprocity or correspondence to be 
found between the various parts of the 
figures. ' 

The ordinates of the equilibrium poly- 
gon (vertical distances) correspond to the 
segments of the resolving line (horizontal 
forces), each of these being proportional 
to the bending moments of a simple 
girder sustaining the given weights, and 
resting without constraint upon supports 
at its two extremities. 

The segments of the resultant line 
(vertical distances) correspond to the 
abscissas of the equilibrating polygon 
(horizontal forces) each of these being 
proportional to the bending moments of 



82 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



a simple girder sustaining ttie given 
weights and resting without constraint 
upon a support at their center of gravity. 

The segments of any pseudo resultant 
line, parallel to the resultant, which are 
cut off by the sides of the equilibrium 
polygon, are proportional to the bending 
moments of a girder supporting the 
given weights and rigidly built in and 
supported at the point where the line in- 
tersects the girder; to these segments 
correspond the abscissas between the 
equilibrating polygon and a pseudo side 
of it parallel to the pseudo resultant ray. 

The two different kinds of support 
which Ave have supposed, viz. support 
without constraint and support with con- 
straint, can be treated in a somewhat 
more general manner, as appears when 
we consider that at any point of support 
there may be, besides the reaction of the 
support, a bending moment, such as 
would be induced, for instance, when 
the span in question forms part of a con- 
tinuous girder, or when it is fixed at the 
support in a particular direction. In 
such a case the closing line of the equili- 
brium polygon is said to be moved to a 
new position. It seems better to call it 
in its new position a pseudo closing line. 
The ordinates between the pseudo closing 
line and the equilibrium polygon are 
proportional to the bending moments of 



the girder, so supported. It is possible 
to induce such a moment at one point of 
support as to entirely remove the weight 
from the other, and cause it to exert no 
reaction whatever; and any intermediate 
case may occur in which the total weight 
in the span is divided between the sup- 
ports in any manner whatever. When 
the weight is entirely supported at h^ 
then y^e^ is the pseudo closing line of the 
polygon ee. In that case xx becomes the 
pseudo resolving line, and in general the 
ordinates between the pseudo closing 
line and the equilibrium polygon corre- 
spond to the segments of the pseudo 
resolving line, and are proportional to 
the bending moments of the girder. 
This general case is not represented in 
Figs. 8 and 9; but the particular case 
shown, in which the total weight is 
borne by the left pier, gives the equa- 
tions 

^3/3=«^i«'2» ^4/4='^,a'3» ^J,='^r^.^ etc. 

In order to represent the general case 
in which the weights, supported by the 
piers, are not the same as in the case of 
the simple girder, by reason of some kind 
of constraint, we propose to treat the case 
of the straight girder, fixed horizontally 
at its extremities; but it is necessary 
first to discuss the following auxiliary 
construction. 



SUMMATION POLYQON. 
Fig. 10 




THE SUMMATION POLYGON. 

In Fig. 10 let aabh be any closed 



figure of which we wish to determine the 
area. The example which we have 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



83 



chosen is that of an indicator card taken 
from page 12 of Porter's Treatise on 
Richard's Steam Indicator, it being a 
card taken from the cylinder of an old- 
fashioned paddle-wheel Cimarder, the 
Africa. The scale is such that ap^ is 
26.9 pounds per square inch and 06 
parallel to the atmospheric line is the 
length of the stroke. 

Divide the figure by parallel lines afi^^ 
(ifi^, etc. into a series of bands which 
are approximately trapezoidal. A suffi- 
cient number of divisions will cause this 
approximation to be as close as may be 
■desired. The upper and lower bands 
may in the present case be taken as ap- 
proximating sufficiently to parabolic 
areas. Let 06 be perpendicular to cifi^^ 
etc., then will 01, 12, etc., be the heights 
of the partial areas. Lay off 

KK=i{^A-^^A)y etc. 
then will these distances be the bases of 
the partial areas. Assume any point c 
at a distance I from 06 as the common 
point of the rays of a pencil passing- 
through 0,. 1, 2, etc.; and draw the 
parallels hs : then from any point v^ of 
the first of these make v^s^ \\ cO, and 
s^s^ II cl, s^s^ II c2, etc. 

The polygon ss is called the summa- 
tion polygon, and has the following 
properties. 

By similarity of triangles 

; : 01 : : hA- v,v^, .'. 0\.hA=l'V,v^ 

is the area of the upper band. Similarly 
Vi.h^h^=^l.v^v^ is the area of the next 
band, and finally 

OQS{hA=l^v,v=lp 

is the total area of the figure. 

In the present instance we have taken 
/=06, the length of stroke, conse- 
quently/) is the average pressure during 
the stroke of the piston, and is 21.25 
pounds, which multiplied by the volume 
of the cylinder gives the work per stroke. 

This method of summation, which ob- 
tains directly the height 2^ of a rectangle 
of given base I equivalent in area to any 
given figure, is due to Culmann, and is 
applicable to all problems in planimetry; 
it is especially convenient in treating the 
problems met with in equalizing the 
areas of profiles of excavation and em- 
bankment, and is frequently of use in 



dividing land. It is much more ex- 
peditious in application than the 
method of triangles founded on Euclid, 
and is also, in general, superior to 
the method of equidistant ordinates, 
whether the partial areas are then 
computed as trapezoids or by Simp- 
son's Rule; for it reduces the number 
of ordinates and permits them to be 
placed at such points as to make the 
bands approximate much more closely 
to true trapezoids than does the method 
of equidistant ordinates. 

GIRDEK WITH FIXED ENDS. 

It is to be understood that by a girder 
with fixed ends, we mean one from which 
if the loading were entirely removed, 
without removing the constraint at its 
ends, there would be no bending moment 
at any point of it, and, when the loading 
is applied to it the supports constrain 
the extremities to maintain their original 
direction unchanged, but furnish no 
horizontal refsistance. Under those cir- 
cumstances the girder may not be 
straight, and may not have its supports 
on the same level, but it will be more 
convenient to think of the girder as 
straight and level, as the moments, etc., 
are the same in both cases. 

Suppose in Fig. 11 that any weights 
lo^w^, etc. are applied at A^, /^3, A^, h^, to 
a girder which is supported and fixed 
horizontally at h^ and Ag. With p as the 
pole of a force pencil draw the equili- 
brium polygon ee as in Fig. 8. The re- 
sultant passes through m. 

It is shown in my New Constructions 
in Graphical Statics, Chapter II, that the 
position of the pseudo closing line k'7c\ 
in case the girder has its ends fixed as 
above stated, is determined from the 
conditions that it shall cut the curve ee 
in such a way that the moment area 
above Jc'Jc' shall be equal to that below 
]c'k\ and also in such a way that the 
center of gravity of the new moment 
area shall be in the same vertical as the 
original moment area. 
, To find the center of gravity of the 
moment area ek\ determine the areas of 
the various trapezoids of which it is com- 
posed by help of the summation poly- 
gon ss. In constructing ss we make 
hj.—k^e^, h^'l=k^e^-\:k^e^, etc., and using 
V as the common point of the pencil we 



84 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 




shall have h^v.h^z^ = twice the area of 
the moment area. We have used the 
sum of the two parallel sides of each 
trapezoid instead of half tjiat quantity 
for greater accuracy. 

Now lay off from 2^, z^z^-=h^z^^ 
z^z^=.h^z^, etc., as a weight line and 
assume the pole jo'. 

Of the triangle hji^e^^ one-third 
rests at li^ and two-thirds at A^; 
make 'z^z^=^\z^z^^ it is the part of the 
area applied at A^. Of the area h^e^eji.^^ 
one half, approximately, rests at h^ and 
one half at A„. Bisect z^z^ at z\ then 



"2 5 



zJz^ 



rests at A„. Bisect each of the 



2* 



Other quantities z^z^^ etc. except z^z^^ in 
which make 2^2/= J z^z^. With the 
weights z'z' so obtained, construct the 
second equilibrium polygon yy, 'which 
shows that the center of gravity of the 
moment area is in the vertical through 
n. There is a balancing of errors in this 
approximation which renders the posi- 
tion of n quite exact; if, however, 
greater precision is desired, determine 
the centers of gravity of the trapezoids 
forming the moment area, and use new 
verticals through them as weight lines, 
with the weights zz instead of the 
weights z'z' . 

Draw A^erticals which divide the span 
into three equal parts, — they cut ny^ and 
ny^ at t^ and ^g, and draw p't' \\ t^t^. 
Then is t^t^nt^t^ an equilibrium polygon 
due to the force z^z^ applied at n, and to 
the forces z^t' ^ and t'z^ applied at t^ and 
^3 respectively. As explained when 



hh'=\ t'z 



treating this matter in the New Con- 
structions in Graphical Statics, z^t' and 
t'z^ are proportional to the bending mo- 
ments at the extremities of the fixed gird- 
er. In this case, since we have taken 
we find that hJc^--rz\zJL'^ and 
are the end moments, and 
they fix the position of the pseudo clos- 
ing line. Draw pq' \\ 7c' k' then are w^q'^ 
and q'w^ the reactions of the piers. The 
pseudo resultant is at m'. 

To obtain the same result by 
help of a frame pencil, let Fig. 12 
represent the same weights applied 
in the same manner as in Fig. 11. 
Choose the vertex v, and draw the 
equilibrating polygon dd, etc. as in Fig. 
8. Make \\=r^r^, \2=r^r^ + r^r^, etc., 
since these quantities are proportional 
to the bending moments as previously 
shown. With v as the common point of 
the rays of a pencil, find h^z^ by the help 
of the summation polygon 55 just as in 
Fig. 11. 

Lay off the second weight line z^z^\ 
etc., just as in Fig. 11, and with v as 
vertex construct the second equilibrating 
polygon XX. Then as readily appears 
vn 11 z^x, determines 7i the center of 
gravity of the moment area. Make z^x^ 
II ?;^.^ and x^x^\\vt^\ if t^ and t^ divide 
the span into three equal parts, then the 
horizontal through x^ fixes ^' correspond- 
ing to t' in Fig. 11. 

To find the position of the pseudo 
resolving line and its segments pro- 
portional to the new bending mo- 



A NEW GENEEAL METHOD IN GRAPHICAL STATICS. 



85 



ments, lay off rj=\{t'z^—z^t') the differ- 
ence of the bending moments at the 
ends, and make />/ || r^w^ and prolong 
it^r^ until they meet at r/ which is on the 
pseudo resolving line. Then lay off 
r^r'=\z^t' and ry=\t'z^'- upon this 
pseudo resolving Mne r'q', then r'r^', ^'^g'? 
etc., are the bending moments when the 
girder is fixed at the ends. For by simi- 
larity of triangles 






r 't ' 

16 



9.9.'^ 



is the moment, and qq' is the weight 
which is transferred from one support to 
the other by the constraint, hence r'q' is 
the correct position of the pseudo resolv- 



ing line. Thence follows the proof that 
the bending moments are proportional 
to intercepts upon this line in a manner 
precisely like that employed in Fig. 9. 

Again, draw m^ \\ wy and vi^ \\ uy, 
then are i^ and i^ the points of inflexion 
of the girder when the bending moment 
vanishes, being in reality points of sup- 
port on which the girder could simply 
rest without constraint and have tlie 
pseudo resultant in that case as the true 
resultant. 

In Figs. 11 and 12 we have taken 
11=^ V, consequently the new moments 
can be directly compared, the ordinates 
k'e being equal to the corresponding 
segments rV. 




Apparently in this example Fig. 12 
presents a construction somewhat more 
compact than that of Fig. 11, it is cer- 
tainly equally good. 

It remains to remark before proceed- 
ing to further considerations of a slight- 
ly different character, that we owe to 
the genius of Culmann* the establishment 
of the generality of the method of the 
equilibrium polygon. 

He adopted the funicular polygon, 
some of whose properties had long been 
known, and upon it founded the general 
processes and methods of systematic 
work which are now employed by all. 

Furthermore it should be stated that 
parallelograms of forces were com- 
pounded and applied in such a way as to 

* Graphische Statik. Zurich, 1866. 



give rise to a frame pencil and equili- 
brating polygon by the illustrious 
Poncelet* who by their use determined 
the centers of gravity of portions of the 
stone arch. Whether he recognized 
other properties besides the simple de- 
termination of the resultant of parallel 
forces, I am not informed, as my 
knowledge of Poncelet's memorial is de- 
rived from so much of his work as 
Woodburyf has incorporated in his 
graphical construction for the stone 
arch. 

So far as known, the method has been 
advanced by no one of the numerous 
recent writers upon Graphical Statics 



* Memorial de 1' officier du Genie. No. 12. 
t Treatise on the Stability of the Arch. D. P. Wood- 
bury, New York, 1858. 



86 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



which would certainly have been the 
case had Poncelet established its claim 
to be regarded as a general method. 
I think the method of the frame pen- 
cil may now fairly claim an equal gen- 
erality and importance with that of the 
equilibrium polygon. 

ANY FORCES LYING IN ONE PLANE, AND 
APPLIED AT GIVEN POINTS. 

We have previously referred to this 
problem, having treated a particular case 
of it in Fig. 2 ; and subsequently cer- 
tain statements were made respecting the 
indeterminateness of the process for find- 
ing the reactions of supports in case the 
applied forces were not vertical. 

The case most frequently encountered 
in practice is wind-pressure combined 
with weight, and we can take this case 
as being sufficiently general in its nature; 
so that we are supposed to know the 
precise points of application of each of 
the forces, and its direction. Now it 
may be that the reaction of the supports 
cannot be exactly determined, but in all 
cases an extreme supposition can be made 
which will determine stresses in the 
framework which are on the safe side. 

For example, if it is known that one 
of the reactions must be vertical, or nor- 
mal to the bed plate of a set of support- 
ing rollers, this will fix the direction of 
one reaction and the other may then be 
found by a process, like that employed 
in Fig. 2, of which the steps are as fol- 
lows : 

Resolve each of the forces at its point 
of application into components parallel 
and perpendicular to the known direction 
of the reaction, which we will call verti- 
cal for convenience, since the process is 
the same whatever the direction may be. 
By means of an equilibrium polygon or 
frame pencil find the line of action of 
the resultant of the horizontal compo- 
nents, whose sum is known. Then this 
horizontal resultant, can be treated pre- 
cisely as was the single horizontal force 
in Fig. 2, which will determine the alter- 
ation of the vertical components of the 
reactions due to the couple caused by the 
horizontal components. 

Also, find by an equilibrium polygon, 
or frame pencil, the vertical reactions due 
to the vertical components. Correct the 
point of division q of the weight line as 
found from the vertical components by 



the amount of alteration already found 
to be due to the horizontal components. 
Call this point q\ then the polygon of 
the applied forces must be closed by two 
lines representing the reactions, which 
must meet on a horizontal through q' \ 
but one of them has a, known direction, 
hence the other is completely determined. 

This determination causes the entire 
horizontal component to be included in 
a single one of the reactions, and it is 
usually one of the suppositions to be 
made when it is not known that the reac- 
tion of a support is normal to the plane 
of the bed joint. 

Another supposition in these circum- 
stances is that the horizontal component 
is entirely included in the other reaction; 
and a third supposition is that the hori- 
zontal component is so divided between 
the reactions that they have the same 
direction. These suppositions will usu- 
ally enable us to find the greatest possible 
stress on any given piece of the frame by 
taking that stress for each piece which is 
the greatest of the three. 

In every supposition care must be 
taken to find the alteration of the verti- 
cal components due to the horizontal 
components. This is the point which has 
been usually overlooked heretofore. 

KERNEL, MOMENTS OF RESISTANCE AND 
INERTIA : EQUILIBRIUM POLYGON METHOD. 

The accepted theory respecting the 
flexure of elastic girders assumes that 
the stress induced in ^ny cross section 
by a bending moment increases uniform- 
ly from the neutral axis to the extreme 
fiber. 

The cross section considered, is sup- 
posed to be at right angles to the plane 
of action or solicitation of the bending 
moment, and the line of intersection of 
this plane with that of the cross section 
is called the axis of solicitation of the 
cross section. 

The radius of gyration of the cross 
section about any neutral axis is in the 
direction of the axis of solicitation. 

It is well known that these two axes 
intersect at the center of gravity of the 
cross section, and have directions which 
are conjugate to each other in the ellipse 
which is the locus of the extremities of 
the radii of gyration. 

We shall assume the known relation 



A NEW. GENERAL. METHOD IN GRAPHICAL STATICS. 



87 



in which 31 is the magnitude of the 
bending moment, or moment of resistance 
of the cross section, S is the stress on 
the extreme fiber, /is the moment of in- 
ertia about any neutral axis aj, and y is 
the distance of the extreme fiber in the 
direction of the axis of solicitation, ^. e. 
the distance between the neutral axis x 
and that tangent to the cross section 
which is parallel to x and most remote 
from it, the distance being measured 
along the axis of solicitation. 

Let M=Sm in which m is called 
the "specific moment of resistance" of 
the cross section; it is, in fact, the 
bending moment which will induce a 
stress of unity on the extreme fiber. 

Now I=¥A 

in which k is the radius of gyration and 
A is the area of the cross section. 



Let 



l^-7-y^=.i\ .'. 77i=rA, 



is the specific moment of resistance 
about X, and when the direction of x 
varies, r varies in magnitude: r is called 
the " radius of resistance " of the cross 
section. The locus of the extremity of 
r, taken as a radius vector along the 
axis of solicitation, is called the "ker- 
nel." 

The kernel is usually defined to be the 
locus of the center of action of a stress 
uniformly increasing from the tangent 
to the cross section at the extreme fiber. 
It was first pointed out by Jung,* and 
subsequently by Sayno, that the radius 
vector of the kernel is the radius of 
resistance of the cross section measured 
on the axis of solicitation. This will 
also appear from our construction by a 
method somewhat different from that 
heretofore employed. 

Jung has also proposed to determine 
values of 7c, by first finding r/ and has 
given methods for finding r. We shall 
obtain r by a new method which renders 
the proposal of Jung in the highest 
degree useful. " 

The method heretofore employed by 
Culmann and other investigators has 
been to find values of k first, and then 
having drawn the ellipse of inertia to 



* " Rappresentazioni graflsche dei momenti resistenti 
dj. una sezione plana." G. Jung, Rendiconti dell' Instituto 
Sombardo, Ser. 2, t, IX, 1876, No. XV. " Complemento 
alia nota precedente." No. XVI. 



construct the kernel as the locus of the 
antipole of the tangent at the extreme 
fiber. The method now proposed is the 
reverse of this, as it constructs several 
radii of the kernel first, then the corre- 
sponding radii of gyration, and from 
them the ellipse, and finally completes 
the kernel. In the old process there are 
inconvenient restrictions in the choice of 
pole distances which are entirely avoided 
in the new process. 

Let the cross section treated be that 
of the X I'^il represented in Fig. 13, 
which is 4|x2j inches and ^ inch thick. 
We have selected a rail of uniform 
thickn'ess in order to avoid in this small 
figure the numerous lines needed in the 
summation polygon for determining the 
area; but any cross section can be treat- 
ed with ease by using a summation j^oly- 
gon for finding the area. 

To find the center of gravity, let the 
weights w^io^ and toj.o^, which are propor- 
tional to the areas between the verticals 
at b^b^ and b^b^ be applied at their centers 
of gravity a^ and a^ respectively; then 
the equilibrium polygon c^c^, having the 
pole ^9^, shows that ois the required cen- 
ter of gravity. 

Let the area b^b^ be divided into two 
parts at o, then w^io^ and w^to^ are 
weights proportional to the areas b^o and 
063 respectively; and c^o^c^ is the equili- 
brium polygon for these weights applied 
at their centers of gravity a^ and a^. 

The intercepts mm have been previ- 
ously shown to be proportional to the 
products of the applied weights by their 
distances from the center of gravity o. 

We have heretofore spoken of these 
products as the moments of the weights 
about their common center of gravity 0. 
But the weights in this case are areas 
and the product of an area by a distance 
is a volume. Let us for convenience call 
volumes so generated "stress solids." 
The elementary stress solids obtained by 
multiplying each elementary area by its 
distance from the neutral axis will cor- 
rectly represent the stresses on the dif- 
ferent parts of the cross section, and they 
will be contained between the cross sec- 
tion and a plane intersecting the cross 
section along the neutral axis and mak- 
iug an angle of 45° with the cross sec- 
tion. 

If b^b^ is the ground line, ^^^3 and d^cl^ 
are the traces of the planes between 



S8 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 




which the stress solid lies on a plane at 
right angles to the neutral axis. 

The distances of the centers of gravity 
of the stress solids from o are also the 
distances of the points of application of 
the resultant stresses, and the magnitude 
of the resultant stresses are are propor- 
tional to the stress solids. The stress 
solids may be considered to be some kind 
of homogeneous loading whose weight 
produces the stress upon the cross section. 
The moment of inertia J is the mo- 
ment of this stress with respect to o. 

Now the intercept m^m^ represents 
the weight of the stress solid whose 
profile is ob^d^. Its point of applica- 
tion is ^3, if og^=^ob^. Similarly the 
weight m^m^ has its point of application 
at g^ if og^=-\oh^. And the weight m^rn^ 
is applied in the vertical through g^ ; for 
the profile of this stress solid is the trape- 
zoid hj)^d^d^, and g^ is its center of grav- 
ity found geometrically. In case the 



area is divided into narrow bands paral- 
lel to the neutral axis the points of appli- 
cation coincide sensibly with the centers 
of gravity of the bands. 

Now take any pole p^ and construct a 
second equilibrium polygon ee due to the 
stress solids applied in the verticals 
through g.g^g^. 

The last two sides e^n^ and e^n^ are 
necessarily parallel and have their inter- 
section at infinity, for the total stress i& 
a couple. 

The intercept n^n^ is not drawn through 
the common center of gravity of the 
stress solids, i. e., it is not an intercept 
on the line of the resultant stress, but 
since parallels are everywhere equidis- 
tant this intercept is proportional to the 
moment of the stresses about their center 
of gravity ; in other words n/i^ when 
multiplied successively by the two pole 
distances would be /. We shall not need 
to effect the multiplication. 



A NEW GENEKAL METHOD IN GEAPHICAL STATICS. 



89 



Prolong c^m^ to e^ on the tangent to 
the extreme fiber and draw c^m^W p^ic^, 
then m^m^ represents the product of the 
total weight-area w^w^ by oh^-=-y the dis- 
tance of the extreme fiber, or m^m,^ is 
proportional to the volume of a stress 
solid whose base is the entire cross sec- 
tion and whose altitude is 'b^d^-=^ob^. 

Suppose this stress to be of the same 
sign as that at the right of o, let us com- 
bine it with the stress already treated. 
Its point of application is necessarily at 
0, and its amount is m^m^ if measured 
on the same scale as the other stresses. 

Draw n^e^ lli^a^o? ^^^^ ^^ ^i ^^ ^^ verti- 
cal through e^ the point of application of 
the combined stresses. But the com- 
bined stresses amount to a stress whose 
profile is included between d^d^ and a 
horizontal line through d^^ i.e. to a stress 
uniformly increasing from h^ to h^\ hence 
h^ is a point of the kernel as usually de- 
fined. 

If c^m^ be prolonged to c^ and we draw 
^6^6 lli^i^^i, t-hen m^^ (not shown) is the 
weight of a stress solid of a uniform 
depth h^d^ over the entire cross section; 
and if we draw n^e^ Wp^'n^i then will h^ 
on the vertical through e^ be also in like 
manner a point of the kernel, i.e. the 
point of application of a stress uniformly 
increasing from ^3 to 1)^. 

But now let us examine our construc- 
tion further in order to gain a more 
exact understanding of what the dis- 
tances r^-zoJc^ and r^=zo7c^ signify. 

We have shown that m^m^ represents 
the product of the area of the cross sec- 
tion by the distance ob^ of the extreme 
fiber, i.e. the quantity Ay^\ but 7i^n^ rep- 
resents the moment of this weight when 
applied at ^„ i.e. the product Ay/^. 
Also as previously shown n/i^ repre- 
sented I on the same scale, hence 

I=Ay^r^, but I=Ak^' .\ r=Jc^'-^y^ 

and r^ is the radius of resistance pre- 
viously mentioned. 

In order to determine the radius of 
gyration \, which is a mean proportional 
between r^ and y^, describe a circle on 
b^k^ as a diameter intersecting 771771 at A 
then oh=k^ the semi-axis of the ellipse 
of inertia conjugate to mm as a neutral 
axis. The accuracy of the construction 
is tested by using b^k^ as a diameter and 
finding the mean proportional between 
ok^ and ob^. It should give the same 



result as that just obtained. In our Fig. 
both circles intersect at h. 

It is known from the symmetry of 
figure of the cross section that k^ is one 
of the principal axes. 

In similar manner we construct the 
radius of resistance, etc., when b^b^ is 
taken as the neutral axis. 

Knowing before hand that this line 
passes through the centre of gravity, 
we have taken the weights of the area 
above it in two parts, viz.: that extend- 
ing from 5j^25 ^^^ t^^t from ^^^g, and 
we have taken w/w/ and w^w^ respec- 
tively, as the weights of these. Choose 
any pole jo/ and draw the equilibrium 
polygon c'c'\ use its intercepts m'm\ 
which represent the weights of stress 
solids, as weights and with any pole p^ 
construct the second equilibrium polygon 
e'e' on the verticals through the points of 
application of the stresses. Also find 
TYilm^ the product of the total area by 
the distance of the extreme fiber and 
make n^e^ ||p/m/; then is h^ which is 
on the same vertical as e/ a point of the 
kernel, and ok^^^r^ the radius of resist- 
ance. Use h^b^ as a diameter, then is 
oh'='k^' the radius of gyration, for 

With these two principal axes tlius 
determined, it is possible at once to con- 
struct the ellipse of inertia. In any case 
it will be possible to determine the direc- 
tion of the axis of solicitation correspond- 
ing to any assumed neutral axis by actual 
construction, it being simply necessary to 
find the line through o upon which lie 
the points of application of the positive 
and negative stresses considered separate- 
ly. These axes being conjugate direc- 
tions in the ellipse of inertia, when we 
have found the radii of resistance in 
those two directions we can at once ob- 
tain the corresponding radii of gyration 
which are conjugate semi-diameters, and 
so draw the ellipse. 

After the ellipse is drawn the kernel 
can be readily completed by making r 
in every direction a third proportional to 
the distance of the extreme fiber and 
the radius of gyration. 

We are assisted in drawing the kernel 
by noticing that to each straight side of 
the cross section there corresponds a 
single point in the kernel, and to each 
non re-entrant angular point aside of the 
kernel, these standing in the mutual re- 



90 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



lation of polar and anti-pole withi respect 
to the ellipse of inertia, as shown by the 
equation Jc^^ry. 

In Fig. 13 the point Ic^ corresponds to 
the left hand vertical side, the point k^ 
to the right hand vertical side, and the 
sides ^\A;/, hjc.^ to the angular points at 
the upper and lower extremities of the 
left side respectively, while the points 



Jc^'k^' at the very obtuse angular points 
of the kernel correspond to the upper 
and lower horizontal sides of the flange. 
The two remaining angular points of the 
kernel correspond to tangent lines when 
they just touch the corners of the flange 
and web, while the intermediate sides 
correspond to the angles at the extremi- 
ties of these lines. 



My Uji 



Ms m Us) Us U2 




kernel, moments of resistance and 
inertia: frame pencil method. 

Let the cross section treated be that 
shown in Fig. 14, which is nearly that 
of a 56 lb. steel rail, the difference con- 
sisting only in a slight rounding at the 
.angles. 

Let the cross section be divided by 
lines perpendicular to the axis of symme- 
try bh at ^2) ^3) Gtc, then the partial areas 
and the total area may be found by a 
summation polygon. 

Take c as the common point of the 



rays through bj)^, etc., and make 01, 02, 
etc., proportional to the mean ordinates 
of the areas standing on the bases bfi^^ 
b^b^, etc. respectively. Draw s^u^^ || cb^y 
s^u^ II 0^2, etc., then will the segments of 
the line ww represent the respective par- 
tial areas, and u^ii^ will represent the 
total area. 

Divide the vertical line %ow into seg- 
ments equal to those of the line nn, then 
is WW the weight line for finding the 
center of gravity, etc., of the cross sec- 
tion. Let a^, «2) <^3) 6tc., be the centers 



A NEW GENERAL METHOD IN GKAPHICAL STATICS. 



91 



of gravity of the partial areas, and let 
V be the vertex of a frame pencil whose 
rays pass through these centers of 
gravity. Draw the equilibrating poly- 
gon dd with its sides parallel to the rays 
of this frame pencil, then the ray vo 
parallel to the closing side yy of the 
equilibrating polygon determines the 
center of gravity o of the cross section, 
according to principles previously ex- 
plained. 

It will be convenient to divide the 
cross section into two parts by the verti- 
cal line o^, which we shall take as the 
neutral axis. The partial areas d^o and 



od, have a/ and 



a„ 



as their centers of 



gravity. Make s^u^ \\ co, then ^o^ which 
corresponds to ii^, divides the weight 
line into two parts, representing the 
areas each side of the. neutral axis, and 
the polygon dd can be completed by 
drawing d^d^ \\ va^' and d^d^ || vaj\ It 
has been previously shown that the 
abscissas yd represent the sum of the 
products of the weights {i.e. areas) by 
their distances from o; and any single 
product is the difference of two success- 
ive abscissas. Project the lengths yd 
upon the horizontal zz by lines parallel 
to yy, then the segments of zz represent 
the products just mentioned. But these 
products are the stress solids or resultant 
stresses before mentioned. Hence zz is 
to be used as a weight line and is trans- 
ferred to a vertical position at the left 
of the Fig. The points of application of 
the resultant stresses may without sensi- 
ble error be taken at the centers of 
gravity a^a^, etc., of the partial areas ex- 
cept in case of the segments of the web 
on each side of o. For these, let oyj 
=§ob^, and og^"=^ob^, then y/ and ^3" 
are the required points of application. 

Now with the weight line zz, which 
consists partly of negative loads, and 
with the same vertex v construct the 
second equilibrating polygon ff, then 
^ifi I'epresents the moment of inertia of 
the cross section, it being proportional 
the moment of the resultant stresses 
about 0. It is seen that the sides f^f^ 
and f^/^ are so short that any small de- 
viation in their directions would not 
greatly affect the result, and that there 
would therefore have been little error if 
the resultant stresses in the web had 
been applied at a/ and a^'\ 

Again, draw dd^^ || vb^, then the hori- 



zontal line dw^ [=d^d') represents Ay^, 
the product of the total weight w^w^ 
{i. e. the total area of the cross sec- 
tion), by the distance of the extreme 
fiber ob^-=y^. Use this as a stress solid 
or resultant stress applied at o and hav- 
ing a weight zz=d^d\ and draw oj \\ zf ^, 
j being at the same vertical distance from 
bb as V is; then is 7*;^, which on Ihe same 
vertical at 7, a point of the kernel. For 
Jc^ is such a point that the product of oJc^ 
(=rj by the weight zzJ^=Ay;) is zj ^-1 
on the same scale as Zwas previously 
measured. 

Similarly draw w^d^ \\ vb^ and make 
z^z^=d^d^\ also draw iA;^ i|//g: then is 
Jc^ another point of the kernel as appears 
from reasons like those just given in 
case of k^. 

Use bjc^ as a diameter, then oh is a 
semi-axis of the ellipse of inertia. The 
same point h should be found by using 
kjb^ as a diameter. Another semi-axis 
of the ellipse of inertia with reference 
to bb as a neutral axis, and conjugate to 
oh can be determined, using the same 
partial areas, by finding the centers of 
gravity and points of application of the 
stresses of the partial areas on one side 
of bb, the^process being similar to that 
employed in Fig. 13, except in the em- 
ployment of the frame pencil instead of 
the equilibrium polygon. 

It is to be noticed that the closing side 
f^z^ of the second equilibrating polygon 
ff is parallel to a resultant ray which 
intersects bb at infinity, the point of ap- 
plication of the resultant of the applied 
stresses, i. e. the stresses form a couple. 

When the ellipse of inertia has been 
found by determining the magnitude and 
direction of two conjugate axes, the ker- 
nel can be readily completed as has been 
shown in connection with Fig. 13. 

trJSriFOEMLY VARYIISrG STRESS IJST GENERAL. 

The methods employed in Figs. 13 
and 14 are applicable also to any uni- 
formly varymg stress, for a stress which 
uniformly increases from any neutral 
axis X through the center of gravity of 
the cross section can be changed into a 
stress which uniformly increases from 
same parallel axis x' at a distance y 
from X by simply combining with the 
former a stress uniformly distributed 
over the cross-section and of such intens- 



92 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



ity as to make the resultant Id tensity 
zero along x' . 

In the construction given in Figs. 13 
and 14 it is only necessary to use the 
proposed line x' at a distance y^ from o, 
instead of the tangent to the extreme 
fiber at a distance y^ or y^ from o, when 
we wish to determine the weight or 
volume of the resultant stress solid, its 
moment about o, and its center of gravi- 
ty or application. 

Since the locus of the center of appli- 
cation of the resultant stress is the anti- 
pole of x' with respect to the ellipse of 
inertia, it is evident that when the pro- 
posed axis x^ lies partly within the cross 
section the center of application of the 
resultant stress is without the kernel, 
and that when x' is entirely without the 
cross section its center of application is 
within the kernel. 

It is frequently more convenient to 
determine the center of application from 
the kernel itself than from the ellipse 
of inertia. This can be readily found 
from the equation which we are now to 
state 

in which equation Ay^ and Ay^ are the 
volumes of the stress solids which if 
uniformly distributed and compounded 
with the stress whose neutral axis is a;, 
will cause the resultant stresses to vanish 
at distances y^ and y^, respectively; 
while T^ and r^ are the distances from o 
of the respective centers of application 
of these stresses. 

The truth of the equation is evident 
from the fact that the moment about o 
of any stress solid uniformly distributed 
is zero, hence the composition of such a 
stress with that previously acting will 
leave its moment unchanged. 

From the equation just stated we 
have , 

from which r^ can be found by an ele- 
mentary construction, since y^^ y^ and r^ 
are known quantities. When it is de- 
sired to express these results in terms of 
the intensities of the actual stresses, 

let p^-=ny^ be the mean stress; 
and let p^'=n [y^-^-y^ be the greatest, 
and let p^=-n {y^—y^ be the least 
intensity at the extreme fiber: 



then ny^ =p^ ' —ny^ =p^' -p^ 



or 



or 



Po 'P/-Po ' :^ :^o 



Po 'Po-P, 



: r : T 



in which r^ and r^ are the two radii of 
the kernel. 

DISTRIBUTION OF SHEARING STRESS. 

It is well known that the equation 
dM^= TdZy expresses the relation of the 
total shearing stress T sustained at any 
cross section of a girder to the variation 
dM of the bending moment il/ at a 
parallel cross-section situated at the 
small distance dz from the first men- 
tioned cross section. 

We have already treated the normal 
components of the stress caused by the 
bending moment M'. we shall now treat 
the tangential component or shear which 
accompanies any variation of the bend- 
ing moment. 

We shall assume as already proved 
the following equation* which expresses 
the intensity q of the shearing stress at 
any point of the cross section:'- 

Iqx- TV 

in which x is the width of the girder 
measured parallel to the neutral axis at 
any distance y from the neutral axis, and 
q is the intensity of the shearing stress 
at the same distance, I'\% the moment of 
inertia of the cross section about the 
neutral axis, T is the total shear at this 
cross section, and V is the volume of 
that part of one of the stress solids used 
in finding the moment of inertia which 
is situated at a greater distance than y 
from the neutral axis, i.e. in Fig. 13 if 
we were finding the value of q at h^, 
with respect to om^ as the neutral axis, 
then y would signify the stress solid 
whose profile is d^d^ ^fix- -^^j however, 
makes no difference whether we define V 
as the stress solid situated at the left or 
at the right of h^\ for, since the total 
stress solid, positive and negative, is 
zero, that on either side of any assumed 
plane is the same. 

The first step in our process is to find 
the intensity of the shear at the neutral 
axis, which we denote by q^\ and if we 
also call x^ the width here and V^ the 
volume of either of the two equal stress 

• See Rankine's Applied Mechanics. Eighth Edition, 
Art. 309, p. 338. 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



93 



solids between this axis and the extreme 
€ber, we have 

when d is the distance between the cen- 
ters of application of the equal stress 
solids, i.e., d is the arm of the couple of 
the resultant stresses. Also T=Aq 
when A is the total area of the cross 
section and q is the mean intensity of 
the shearing stress. Hence at the neu- 
tral axis we have the equation 

q^x^d=Aq=T 



Now the length of the arm d is found 
in Fig. 13 by prolonging the middle side 
{i.e. the side through w,) of the second 
equilibrium polygon until it intersects 
the first side and the last. These inter- 
sections will give the position of the 
centers of gravity of the stress solids on 
either side of o. 

In Fig. 14 the same points are found 
by drawing rays from v parallel respect- 
ively to zj^ and /y„ until they inter- 
sect aa. 

In Fig. 15 the points /, and f^ are 
found by either of these methods and 
fxfi=^^ is the required distance. 




Now in Fig. 15 let the segments uxi 
of the summation polygon be obtained 
just as in Fig. 14, and parallel to uu 
draw a line through s representing the 
width of the cross section x^ on the same 
scale as before used in constructing the 
summation polygon. Also make su^ \\ 
cf^^ and su || c/*^, c being the common 
point in the rays of the pencil of the 
summation polygon for finding the area. 

Then uu^ represents the product x^d 
on same scale that u^u^ represents A. 



Now draw from any point ^ rays to w^, 
li and 2^^,_and also a parallel to* iii^ at a 
distance q and intersecting iu at some 
point t^ such that tt^^jf to such a scale 
as may be convenient. The mean intens- 
ity q is supposed to be a known quanti- 
ty, and U^ II uu. Then from the proposed 
equation we have the proportion 



x^d 



A 



or 



uu„ 



u^u^ 



<1 



2o 



tt. 



94 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



Hence tt^ represents the intensity of the 
shearing stress at the neutral axis on 
the same scale that tt^ represents the 
mean intensity. 

This first step of our process has de- 
termined the intensity of the stress at 
the neutral axis relatively to the mean 
stress; the second step will determine 
the intensity of the stress at any other 
point relatively to the stress at the neu- 
tral axis. When this last point is all 
that is desired the first step may be 
omitted. 

The equation Ixq= TV may be written 
xq^=-c F, in which c= T-^ F is a constant. 
At the neutral axis this equation is 

i»o^o=^^^oOi' ^0 : ^0 — a;„ • ^ 
In Fig. 15 lay off the segments of the 
line zz just as in Fig. 14; then z^z^ rep- 
resents the weight or volume F^; also 
make £cO, cc2, x6, etc., proportional to 
width of the girder at o, b^, b^^ etc., and 
lay off z^r=z^rJ = U^, 

Draw pO \\ i\z^^ then by similar tri- 
angles 

^1^0 * ^/o * •' ^0 : xp 
or V^\q^'.\x^\G 

.*. px represents the constant c. 

Now the several segments z^z^^ z^z^^ z^z^^ 
etc., represent respectively the values of 
F^, Fg, F4, or the stress solids between 
one extreme fiber and b^, b^, b^, etc.; it 
is of no consequence which extreme fiber 
is taken as the stress solid is the same 
in either case. 

Now using 2^ as a pole draw rays to 
2 3 4 5 etc., and make s./^ II />2, z^r^ 11^93, 
etc., then by similar triangles 

v., ' z,r^ : : a;2 : c, or x^q^z=cV^ 

and z.z^ : z^i\ : : a;3 : c, or x^q^'^^cV^ 

etc., etc., and z^i\^ z^r^, etc., represent 
the intensity of the shearing stresses at 
^2, ^3, etc. These can be constructed 
equally well by drawing rays from z^ 
parallel to the rays at p, from which we 
obtain 

V/=2i^j 23^'=2^i^5 etc. 

Now lay off b^y^-z^r^, ^,y,=^,'^z, etc., 
then the ordinates by of the polygon yy 
represent the intensity of the shearing 
stress on the same scale that tt^-=z^r^ rep- 
resents the intensity q^ at the neutral 
axis, and on the same scale that U^-=oy' 

represents the mean intensity q. The 



lines joining y^^ 2/3, etc., should be 
slightly curved, but when they are 
straight the representation is quite 
exact. 

RELATIVE STRESSES. 

It is proposed here to develop a new 
construction which will exhibit the rela- 
tive magnitude of the normal compo- 
nents of the stresses produced by a 
given system of loading in the various 
cross-sections of a girder having a varia- 
ble cross section. The value of such a 
construction is evident, as it shows 
graphically the weakest section, and in- 
vestigates the fitness of the assumed dis- 
position of the material for sustaining 
the given system of loading. 

The constructions heretofore given 
for the kernel and moments of resistance 
at any given cross section admit of the 
immediate comparison of the normal 
components of the stresses produced in 
that single cross section when different 
neutral axes are assumed, but by this 
proposed construction, a comparison is 
effected between these stresses at any 
different cross sections of the same gird- 
er or truss. 

In the equation previously used 

M^ SI-^y= SAk'-^y= SAr 

in which 3f is the moment of flexure 
which produces the stress S in the ex- 
treme fiber of a cross section whose area 
is A and whose radius of resistance is r, 
we see, since the specific moment of re- 
sistance m:=Ar is the product of two 
factors, that the same product can result 
from other and very different factors. 

For example, let 971= Ay in which A^ 
is the area of some cross section which 
is assumed as the standard of comparison, 
and ')''=Ar-t-A^^ = ar, when a=^A-i-A^, 
Then is Ay the specific moment of re- 
sistance of a cross section of an assumed 
area A^ which has a. different disposition 
of material from that whose specific 
moment of resistance is Ar, but the 
cross sections A and A^ are equivalent 
to each other in this sense, that they 
have the same specific resistance, and 
consequently the same bending moment 
will produce equal stresses in the 
extreme fiber in each. 

The two cross sections do not have 
the same moment of inertia, and so the 
deflections of the girder would be 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



95 



changed by substituting one cross sec- 
tion for the other. We shall then speak 
of them as equivalent only in the former 
sense, and on the basis of this definition, 
state the result at which we have 
arrived thus: Equivalent cross sections 
under the action of the same bending 
moment, have the same stresses at the 
extreme fiber (though they are not 
equally stiff) ; hence in comparing 
stresses equivalent cross sections may be 
substituted for each other (but they may 
not be so substituted in comparing de- 
flections). 

It is proposed to utilize this result by 
substituting for any girder or truss hav- 
ing a variable cross section A or a varia- 
ble specific moment of resistance wliose 
magnitude is expressed by the variable 
quantity Ar^ a, different one having a 
cross section everywhere of constant 



area A^, but of such disposition of mate- 
rial that its specific moment of resistance 
is Ay = Ar at corresponding cross sec- 
tions. 

The proposed substitution is especially 
easy in case of a truss, for in it the value 
of r varies almost exactly as its depth, 
as may be seen when we compute the 
value of m=Ak^-T-y=Ar 
in this case. 

Since the material which resists 
bending is situated in the chords alone 
and is all approximately at the same 
distance from the neutral axis we have 
k^=3/=^r=ih very nearly when h is the 
distance between the chords, .*. tnr=^Ah 
nearly. Even when the two chords are 
of unequal cross section and tlie neutral 
axis not midway between them the same 
result holds when the ratio of the two 
cross sections is constant. 




In Fig. 16 let axe be the axis of a gird- 
er sustaining at the points iCj, ic^, etc., 
the weights c^c^, c^Cg, etc. Lay off the 
ordinates xy at each of the points at 
which weights are applied, so that xy = 
At on some assumed scale: then since 



Ay^=Ar—xy, xy varies as r\ the radius 
of resistance of a girder having at every 
point a cross section A^ so disposed as 
to be equivalent to that of the given 
girder xx. 

Assume some form of framing con- 



96 



A NEW GENERAL METHOD IN GRAPHICAL STATICS. 



necting the points xy as shown in the 
Fig., and suppose the weights applied 
at the points yy of the lower chord, the 
points of support being at y^ and y^. 
Then by a method like that employed in 
Fig. 3, we obtain the total stresses ea^^ 
ea^, ea^, etc., in the segments of the 
upper chord which are opposite to 2/j, y^, 
2/3, etc. Now these total stresses are 
resisted by a cross section of constant 
area A^^ consequently they have the 
same ratio to one another as the intensi- 
ties per square unit; or further, they 
represent, as we have just shown, the 
relative intensities of the stresses on the 
extreme fiber of the given girder. 

It is well known from mechanical 
considerations, that the stress, in the 
several segments of the upper chord is 



dependent upon the loading and upon 
the position of y^, y^? ^tc, and is not 
dependent upon the position of the 
joints in the upper chord. Of this fact 
we offer the following geometrical proof 
derived from the known relations be- 
tween the frame and force polygons. 

We know, if any joint of the upper 
chord, such as eafi^ for example, be re- 
moved to a new position, such as v, that 
so long as the weights c^c^, c^Cg, etc, are 
unchanged, that the vertex 6, of the tri- 
angle eap^ in the force polygon must be 
found on the force line c^/^ || y^y^. We 
shall show that while the side ea^ is un- 
changed, the locus of h^ is the force line 
o^f^\ hence conversely, so long as e^f^ is 
the locus of Z>j, ea^ is unchanged, since 
there can be but one such triangle. 




In Fig. 17 let the two triangles abe^ hnk^ 
have the sides meeting at h and n 
mutually parallel. Let the bases ae and 
hh be invariable but let the vertex h be 
removed to any point d such that hd \\ lik^ 
then will the vertex n be removed to a 
point m such that mn \\ ae. 

For, prolong ad and eb^ and draw 
bf\\ed and dc\\ah^ then is ahfcdea a 
hexagon inscribed in the conic section 
consisting of the two lines af and ec, 
hemce by Pascal's Theorem, the oppo- 
site diagonals ea and c/* intersect on the 
same line as the remaining pairs of oppo- 
site diagonals, ah\\dc and ed \\ hf. But 
this line is at infinity, hence cf\\ ae. 
Also c'f II c/", from elementary considera- 
tions; and c'f II mn from similarity of 



figures, hence mn \\ ae. There are two 
cases, according as mn is above or below 
hhy but we have proved them both. 

Now in Fig. 16 let all the joints in the 
upper chord be removed to w, then the 
segments ea^, (^^cl^-, etc., are unchanged, 
hence ea^, ea^, etc. are unchanged, and 
the assumed framing reduces to the 
frame pencil whose vertex is v. The 
corresponding force polygon is the 
equilibrating polygon dd. 

Hence the frame pencil can be used as 
the assumed framing just as well as any 
other form of framing, and it is unneces- 
sary to use any construction except that 
of the frame pencil and equilibrating 
polygon for finding the relative stresses 
ea^j ^^3) etc. 



A NEW GENERA.L METHOD IN GRAPHICAL STATICS. 



97 



STRESSES IN A HORIZONTAL CHORD. 

If Fig. 16 be regarded as representing 
an actual bridge truss, wliose chords are 
not of uniform cross section; it is seen 
that the total stresses on the horizontal 
chord are given by the segments ea^, ea^, 
etc., which are found from the equili- 
brating polygon alone without regard to 
the kind of bracing in the truss, which it 
is unnecessary to consider; and tliis 
method can be used to take the place of 
that given in connection with Fig. 3 for 
finding the maximum stresses on the 
chords. 

The equilibrating polygon/*/* was con- 
structed to determine the reactions of 
the piers by finding the point e. The 
outer sides of the polygon /'/ intersect 
at g which determines e as explained in 
Fig. 7 in a manner different from that 
given in Fioj. 3. 



This construction sheds new light 
upon the significance of the frame pencil 
and equilibrating polygon. The frame 
pencil is the limiting case of a truss 
when the joints along one chord are re- 
moved to a single point, so that each ray 
may be regarded as compounded of a 
tension member and a compression mem- 
ber, having the same direction, e.r/.y the 
tension member of which y^v is com- 
pounded has the stress d/i^^ and the 
compression member the stress d,^a.^^ but 
if the two be combined, the resultant 
tension is d/I.^. 

In case yy is the equilibrium curve 
due to the applied weights, and v falls 
upon the closing line, the force lines cd 
meet at the pole and the lines ed^, ed^, 
coincide with aa, so that the polygon dd 
is at the pole and infinitely small, and 
the stress in every segment of the upper 
chord is equal to the pole distance de. 



NOTE A. 



ADDENDUM TO PAGE 12, CHAPTER I 



The truth of Proposition IV is, perhaps, not 
sufficiently established in the demonstration 
heretofore given. As it is a fundamental pro- 
position in tlie graphical treatment of arches, 
and as it is desirable that no doubt exist as to 
its validity, we now offer a second proof of it, 
which, it is thought, avoids the difficulties of 
the former demonstration. 

Prop. IV. If in any arch that equilibrium 
polygon (due to the weights), be constructed 
which has the same horizontal thrust as the 
arch actually exerts ; and if it^ closing line be 
drawn from considerations of the conditions 
imposed by the supports, etc. ; and if, further- 
more, the curve of the arch itself be regarded 
as another equilibrium polygon due to some 
system of loading not given, and its closing 
line be also found from the same considera- 
tions respecting supports, etc. ; then when 
these two polygons are so placed that their 
closing lines coincide, and their areas partially 
cover each other, the ordinates intercepted be- 
tween these two polygons are proportional to 
the real bending moments acting in the arch. 

The bending moments at every point of an 
arch are due to the applied forces and to the 
shape of the arch itself. 

The applied forces are these : the vertical 
forces, which comprise the loading and the 
vertical reactions of the piers; the horizontal 
thnist ; and the bending moments at the piers, 
caused by the constraint at these points of sup- 



port. The loading may cause all the other ap- 
plied forces or it may not: in any case the 
bending moments are unaffected by the de- 
pendence or want of dependence of the thrust, 
etc. , upon the loading. 

Now, so far as the loading and the moments 
due to the constraint at the piers are concerned, 
they cause the same bending moments at any 
point of the arch as they would when applied 
to a straight girder of the same span, for 
neither are the forces nor their arms different 
in tbe two cases. 

But the horizontal thrust, which is the 
same at every point ^of the arch, causes a 
bending moment proportional to its arm, 
which is ihe distance of its line of ap- 
plication from the curve of the arch. This 
line of application is known to be the closing 
line; hence the ordinates which represent the 
bending moments due to the horizontal thrust, 
are included between the curve of the arch and 
a closing line drawn in such a manner as to 
fulfill the conditions imposed by the joints or 
kind of support at the piers, hence the curved 
neutral axis of the arch is the equilibrium or 
moment polygon due to the horizontal thrust. 

But the same conditions fix both the closing 
line of the equilibrium polygon which repre- 
sents the bending moments due to the loading 
and to the constraint at the piers, and the clos- 
ing line of the equilibrium polygon due to the 
horizontal thrust. Hence the resultant bend- 
ing moment is found by taking the difference 
of the ordinates at each point, or by laying 
them off from one and the same closing line 
exactly as described in the statement of our 
proposition. 



NOTE B. 



ADDENDUM TO PAGE 10, CHAPTER I 



Attention should be directed to the two 
senses in which M is used in our fundamental 
fojmulae. 

In equation (3) the primary signification of 
M is this : it is the numerical amount of the 
bending moment at the point 0; and if this 
magnitude be laid off as an ordinate, ym is the 
fraction or multiple of it found by equation (8). 

Now M assumes, in the equations (3), (4), (5) 
and (3'), (4'), (5'), a slightly different and sec- 
ondary signification ; viz. , the intensity of the 
bending moment at 0. The intensity of the 
bending moment is the amount distributed 
along a unit in length of a girder, and may be 
exactly obtained as follows : 



/■ 



Mdx, 



^^o{M)=/ 



Mdx. 



In this secondary sense M is geometrically 
represented by an area one unit wide, and hav- 
ing for its height the average value which 
ordinate M, as first found, has along the unit 
considered. 

Thus the M used in the equations of curva- 
ture, bending and deflection is one dimension 
higher than that used in the equation express- 
ing the moment of the applied forces; but the 
double sense need cause no confusion, and is 
well suited to express in the shortest manner 
the quantities dealt with in our investigation. 

Furthermore, in case of an inclined girder 
such as is treated in Prop. V, if the bending 
moment M, which causes the deflection there 
treated, be represented, it must appear as an 
area between two normals to the girder which 
are at the distance of one unit apart. 

In order to apply Prop. V to inclined and 
curved girders, such as constitute the arch, 
with entire exactness, one more proposition is 
needed. 

Prop. If weights be sustained by an in- 
clined girder, and the amount of the deflection 
of this girder, which is caused by the weights, 
be compared with the deflection of an hori- 



zontal girder of the same cross section, and of 
the same horizontal span, and deflected by the 
same weights applied in the same verticals ; 
the vertical component of the deflection of 
the inclined girder, at any point 0, is equal to 
the corresponding vertical deflection of the 
horizontal girder, multiplied by the secant of 
the inclination. 

For the bending moment of both the inclined 
girder and the horizontal girder is the same in 
the same vertical, but the distance along the 
inclined girder exceeds that along the hori- 
zontal girder in the ratio of the secant of the 
inclination to unity; hence the respective mo- 
ment areas have this same ratio ; therefore the 
deflections at right angles to the respective 
girders of their corresponding points are in 
the ratio of the square of the secant to unity: 
and the vertical components of the deflections 
are therefore in the ratio of the secant of the 
inclination to unity. 

In applying this proposition to the graphical 
construction for the arch, it will be necessary 
to increase the ordinate of the moment poly- 
gon at each point by multiplying by the secant 
of the inclination of the arch at that point. 
This is easily effected when the ordinates are 
vertical by drawing normals at each point of 
the arch ; then the distance along the normal 
whose vertical component is the bending mo- 
ment is the value of Mto be used in determin- 
ing the deflection. 

In the arches which we have treated the 
rise is so small a fraction of the span that the 
secant of the inclination at any point does not 
greatly exceed unity ; or, to state it otherwise, 
the length of the arch differs by a compara- 
tively small quantity from the actual span. It 
IS a close approximation under such circum- 
stances to use the moments themselves in de- 
termining the deflections ; and we have so used 
them in our constructions. A more accurate 
result can be obtained by multiplying each 
ordinate by the secant of the inclination of 
the arch at that point to the horizon. 



THE THEORY OF INTERNAL STRESS 



IN 



GRAPHICAL STATICS 



THE THEORY OF INTERNAL STRESS 



IN 



GRAPHICAL STATICS 



u m Q." ■ 



Stress includes all action and reaction 
of bodies and parts of bodies by attrac- 
tion of gravitation, cohesion, electric 
repulsion, contact, etc., viewed espe- 
cially as distributed among the particles 
composing the body or bodies. Since 
action and reaction are necessarily equal, 
stress is included under the head of 
Statics, and it may be defined to be the 
equilibrium of distributed forces. 

Internal stress may be defined as the 
action and reaction of molecular forces. 
Its treatment by analytic methods is 
necessarily encumbered by a mass of 
formulae which is perplexing to any ex- 
cept an expert mathematician. It is 
necessarily so encumbered, because the 
treatment consists in a comparison of 
the stresses acting upon planes in vari- 
ous directions, and such a comparison 
involves transformation of quadratic 
functions of two or three variables, so 
that the final expressions contain such 
a tedious array of direction cosines that 
even the mathematician dislikes to em- 
ploy them. 

Now, since the whole difficulty really 
lies in the unsuitability of Cartesian co- 
ordinates for expressing relations which 
are dependent upon the parallelogram of 
forces, and does not lie in the relations 
themselves, which are quite simple, and, 
which no doubt, can be made to appear 
so in quaternion or other suitable nota- 
tion; it has been tbought by the writer 
that a presentation of the subject from a 
graphical stand point would put the 



entire investigation within the reach of 
any one who might wish to understand 
it, and would also be of assistance to 
those who might wish to read the analyt- 
ic investigation. 

The treatment consists of two princi- 
pal parts: in the first part the inherent 
properties of stress are set forth and 
proved by a general line of reasoning 
which entirely avoids analysis, and 
which, it is hoped, will ijiake them well 
understood; the second part deals with 
the problems which arise in treating 
stress. These problems 'are solved 
graphically, and if analytic expressions 
are given for these solutions, such ex- 
pressions will result from elementary 
considerations appearing in the graphi- 
cal solutions. The constructions by 
which the solutions are obtained are 
many of them taken from the works of 
the late Professor Rankine, who em- 
ployed them principally as illustrations, 
and as auxiliary to his analytic investi- 
gations. 

It is thus proposed to render the 
treatment of stress exclusively graphical, 
and by so doing to add a branch to the 
science of Graphical Statics, which has 
not heretofore been recognized as sus- 
ceptible of graphical treatment. It 
seems unnecessary to add a word as to 
the importance, not to say necessity, to 
the engineer of a knowledge of the 
theory of combined internal stress, since 
all correct designing presupposes such 
knowledge. 



104 



THE THEORY OF INTERNAL STRESS 



SxRiEss ON A Plane. — " If a body be 
conceived to be divided into two parts 
by an ideal plane traversing it In any 
direction, the force exerted between 
those two parts at the plane of division 
is an internal stress?'' — Ranhine. 

A State of Internal Stress is such 
a state that an internal stress is or may 
be exerted upon every plane passing 
through a point at which such a state 
exists. 

It is assumed as a physical axiom that 
the stress upon an ideal plane of divi- 
sion which traverses any given point of 
a body, cannot change suddenly, either 
as to direction or magnitude, while that 
plane is gradually turned in any vvay 
about the given point. It is also as- 
sumed as axiomatic that the stress at 
any point upon a moving plane of divi- 
sion which undergoes no sudden changes 
of motion, cannot change suddenly 
either as to direction or amount. A 
sudden variation can only take place at 
a surface where there is a change of 
material. 

GENERAL PROPERTIES 'OF PLANE STRESS. 

We shall call that stress a plane stress 
which is parallel to a plane; e.g., let the 
plane of the paper be this plane and let 
the stress acting upon every ideal plane 
which is at right angles to the plane of 
the paper be parallel to the plane of the 
paper, then is such a stress a plane 
stress. 

The obliquity of a stress is the angle 
included between the direction of the 
stress and a line perpendicular to the 
ideal plane it acts upon. This last 
plane we shall for brevity call the plane 
of action of the stress, and any line 
perpendicular to it, its normal. In plane 
stress, the planes of action are shown by 
their traces on the plane of the paper, 
and then their normals, as well as their 
directions, the magnitudes of the stresses, 
and their obliquities are correctly rep- 
resented by lines in the plane of the 
paper. 

The definition of stress which has 
been given is equivalent to the state- 
ment that stress \9, force distributed over 
an area in such wise as to be in equili- 
brium. 

In order to measure stress it is neces- 
sary to express its amount per unit of 



area: this is called the intensity of the 
stress. 

Stress, like force, can be resolved into 
components. An oblique stress can be 
resolved into a component perpendicular 
to its plane of action called the normal 
component, and a component along the 
plane called the tangential component or 
shear. 

When the obliquity is zero, the entire 
stress is normal stress, and may be either 
a compression or a tension, i.e., a thrust 
or a pull. When the obliquity is +90% 
the stress consists entirely of a tangen- 
tial stress or shear. If a compression be 
considered as a positive normal stress, it 
is possible to consider a normal tension 
as a stress whose obliquity is +180°, 
and the obliquities of two shears having 
opposite signs, also differ by 180°. 

Fig. 1 




Conjugate Stresses. — If in Fig. 1 
any state of stress whatever exists at o, 
and XX be the direction of the stress on a 
plane of action whose trace is yy, then is 
yy the direction of the stress at o on the 
plane whose trace is xx. Stresses so 
related are said to be conjugate stresses. 

For consider the effect of the stress 
upon a small prism of the body of which 
a^a^a^a^ is a right section. If the stress 
is uniform that acting upon a^a^ is equal 
and opposed to that acting upon a,^a^, 
and therefore the stress upon these 
faces of the prism are a pair of forces in 
equilibrium. Again, the stresses upon 
the four faces form a system of forces 
which are in equilibrium, because the 
prism is unmoved by the forces acting 
upon it. But when a system of forces 
in equilibrium is removed from a sys- 
tem in equilibrium, the remaining forces 
are in equilibrium. Therefore the re- 
moval of the pair of stresses in equili- 
brium acting upon a^a^ and a^a^ from 
the system of stresses acting upon the 
four faces, which are also in equilibrium, 
leaves the stresses upon a^a^ and a^a^ in 
equilibrium. But if the stress is uni- 
form, the stresses on a^a^ and a^a^ must 



IN GRAPHICAL STATICS. 



105 



be parallel to yy, as otherwise a couple 
must result from these equal but not 
directly opposed stresses, which is iu- 
consisLerit with equilibrium. 

This proves the fact of conjugate 
stresses when the state of stress is uni- 
form: in case it varies, the prism, can be 
taken so small that the stress is sensibly 
uniform in the space occupied by it, and 
the proposition is true for varying stress 
in case the prism be indefinitely dimin- 
ished, as may always be done. 




Tangential Stresses. — If in Fig. 2 
the stress at o on the plane xx is in the 
direction xx^ i.e. the stress at o on xx 
consists of a shear only; then there 
necessarily exists some other plane 
througli o, as yy, on which the stress 
consists of a shear only, and the shear 
upon each of the planes xx and yy is of 
the same intensity, but of opposite sign. 

For let a plane which initially coin- 
cides with XX revolve continuously 
through 180° about o, until it again co- 
incides with xx^ the obliquity of the 
stress upon this revolving plane has 
changed gradually during the revolution 
through an angle of 360°, as we shall 
show. 

Since the obliquity is the same in its 
final as in its initial position, the total 
change of obliquity during the revolu- 
tion is 0° or some multiple of 360°. It 
cannot be 0°, for suppose the shear to be 
due to a couple of forces parallel to xx^ 
having a positive moment; then if the 
plane be slightly revolved from its 
initial position in a plus direction, the 
stress upon it has a small normal com- 
ponent which would be of opposite sign, 
if the pair of forces which cause it were 
reversed or changed in sign; or, what is 
equivalent to that, the sign of the small 
normal component would be reversed if 
the plane be slightly revolved from its 
initial position in a minus direction. 
Hence the plane xx^ on which the stress 



is a shear alone, separates those planes 
through o on which the obliquity of the 
stress is greater than 90° from those on 
which it is less than 90°, ^.e., those hav- 
ing a plus normal component from those 
having a minus normal component. 

Since in revolving through -f 1 80° the 
plane must coincide, before it reaches its 
final position, with a plane which has 
made a slight minus rotation, it is evi- 
dent that the sign of the normal com- 
ponent changes at least once during a 
revolution of 180°. But a quantity can 
change sign only at zero or infinity, and 
since an infinite normal component is 
inadmissible, the normal component 
must vanish al, least once during the 
proposed revolution. Hence the obliq- 
uity is changed by 360° or some multi- 
ple of 360° while the plane revolves 180°. 
In fact the normal component vanishes 
but once, and the obliquity changes by 
once 360° only, during the revolution. 

It is not in every state of stress that 
there is a plane on which there is no 
stress except shear, but, as just shown, 
when there is one such plane xx there is 
necessarily another yy, and all planes 
through o and cutting the angles in 
which are h^ and h^ have normal com- 
ponents of opposite sign from planes 
through o and cutting the angles in 
which are h^ and h^. 

To show that the intensity of 
the shear on xx is the same as 
that on yy^ consider a prism one unit 
long and having the indefinitely small 
right section bfiJ)J)^. Let the area of 
its upper or lower face be a^^=hj)^, that 
of its right or left face be a^=bj)^, then 
a^s^ and a^s^ are the total stresses on 
these respective faces if .^^ and s^ are the 
intensities of the respective shears per 
square unit. Let the angle xoy=^i, then 



ttjSj . a^ sm. ^ 



is the moment of the stresses on the 
upper and lower faces of the prism, and 



a/2 • ^1 ^^^' ^ 



is the moment of the stresses on the 
right and left faces; but since the prism 
is unmoved these moments are equal. 



s, ~s„ 



These stresses are at once seen to be 
of opposite sign. 



106 



THE THEORY OF INTERNAL STRESS 





Fig. 3 












i 








X 









4' 






^ 




y 



Tangential Components. — In Fig. 3 
if XX and yy are any two planes at right 
angles to each other, then the intensity 
at o of the tangential component of the 
stress upon the plane xx is necessarily 
the same as that upon the plane yy, but 
these components are of opposite sign. 

For the normal components acting 
upon the opposite faces of a right prism 
are necessarily in equilibrium, and by a 
demonstration precisely like that just 
employed in connection with Fig. 2 it is 
seen that for equilibrium it is necessary 
and sufficient that the intensity of the tan- 
gential component on xx be numerically 
equal to that on yy, but of opposite 
sign. 

State of Stkess. — In a state of plane 
stress, the state at any point, as o, is 
completely defined, so that the intensity 
and obliquity of the stress on any plane 
traversing o can be determined, when 
the intensity and obliquity of the stress 
on any two given planes traversing that 
point are known. 

For suppose in Fig. 4 that the intensi- 
ty and obliquity of the stress on the 
given planes a;a; and yy are known, to 
find that on any plane x'x' draw 
mn II x'x' then the indefinitely small 
prism one unit in length whose right 
section is mno, is held in equilibrium by 
the forces acting upon its three faces. 
The forces acting upon the faces oin and 
on are known in direction from the 
obliquities of the stresses, and, if 2>x and 
Py are the respective intensities of the 
known stresses, then the forces are 
om.px and 07i.py respectively. The re- 
sultant of these forces and the reaction 
which holds it in equilibrium, together 
constitute the stress acting on the face 
mn'. this resultant divided by mn is the 
intensity of the stress on mn and its 



direction is that of the stress on m,n or 
x'x'. 



Fig. 4 




It should be noticed that the stress at 
on two planes as xx and yy cannot be 
assumed at random, for such assumption 
would in general be inconsistent with 
the properties which we have shown 
every state of stress to possess. For in- 
stance we are not at liberty to assume 
the obliquities and intensities of the 
stresses on xx and yy such that when 
we compute these quantities for any 
plane x'x' and another plane y'y' at 
right angles to x'x' in the manner just 
indicated, it shall then appear that the 
tangential components are of unequal 
intensity or of the same sign. Or, again, 
we are not at liberty to so assume these 
stresses as to violate the principle of con- 
jugate stresses. 

But in case the stresses assumed are 
conjugate, or consist of a pair of shears 
of equal intensity and different sign on 
any pair of planes, or in case any stresses 
are assumed on a pair of planes at right 
angles such that their tangential compo- 
nents are of equal intensity but different 
sign, we know that we have made a con- 
sistent assumption and the state of stress 
is possible and completely defined. 

The state of stress is not completely 
defined when the stress upon a single 
plane is known, because there may be 
any amount of simple tension or com- 
pression along that plane added to the 
state of stress without changing either 
the intensity or obliquity of the stress on 
that plane. 

Principal Stresses. — In any state of 
stress there is one pair of conjugate 
stresses at right angles to each other, i.e. 
there are two planes at right angles on 
which the stresses are normal only. 
Stresses so related are said to hQ princi- 
pal stresses. 



IlSr GEAPHICAL STATICS. 



107 



It has been previously shown that if 
a plane be taken in any direction, and 
the direction of the stress acting on it be 
found, then these are the directions of a 
pair of conjugate stresses of which either 
may be taken as the plane of action and 
the other as the direction of the stress 
acting upon it. 

Consider first the case in which the 
state of stress is defined by a pair of 
conjugate stresses of the same sign; i.e., 
the normal components of this pair of 
conjugate stresses are both compressions 
or both tensions. 

It is seen that they are of opposite 
obliquities, and if a plane which initially 
coincides with one of these conjugate 
planes of action be continuously revolved 
until it finally coincides with the other, 
the obliquity must pass through all in- 
termediate values, one of which is 0°, and 
when the obliquity is 0° the tangential 
component of the stress vanishes. But 
as has been previously shown the^'e is 
another plane at right angles to this 
which has the same tangential compo- 
nent; hence the stress is normal on this 
plane also. 

Consider next the case in which the 
pair of conjugate stresses which define 
the state of stress are of opposite sign, 
i.e., the normal component on one plane 
is a compression and that on the other 
a tension. 

In this case there is a plane in some 
intermediate position on which the stress 
is tangential only, for the normal com 
ponent cannot change sign except at 
zero. It has been previously shown that 
in case there is one plane on which the 
stress is a shear only, there is another 
plane also on which the stress is a shear 
only, and that this second shear is of 
equal intensity with the first but of 
opposite sign. Let us consider then that 
the state of stress, in the case we are 
now treating, is defined by these oppo- 
site shears instead of the conjugate 
stresses at first considered. 

Now let a plane which initially coin- 
cides with one of the planes of equal 
shear revolve continuously until it finally 
coincides with the other. The obliquity 
gradually changes from +90° to —90", 
during the revolution, hence at some 
intermediate point the obliquity is 0°; 
and since the tangential component has 
the same intensity on a plane at right 



angles to this, that is another plane on 
which the obliquity of the stress is also 
0°. 

We have now completely established 
the proposition respecting the existence 
of principal stresses which may be 
restated thus: 

Any possible state of stress can be 
completely defined by a pair of normal 
stresses on two planes at right angles to 
each other. 

As to the direction of these principal 
planes and stresses, it is easily seen from 
considerations of symmetry that in case 
the state of stress can be defined by 
equal and opposite shears on a pair of 
planes, that the principal planes bisect 
the angles between the planes of equal 
shear, for there is no reason why they 
should incline more to one than to the 
other. We have before shown that the 
planes of equal shear are planes of 
separation between those whose stresses 
have normal components of opposite 
sign: hence it appears that the principal 
stresses are of opposite sign in any state 
of stress which can be defined by a pair 
of equal and opposite shears on two 
planes. 

It will be hereafter shown how the 
direction and magnitude of the principal 
stresses are related to any pair of con- 
jugate stresses. 

For convenience of notation in discuss- 
ing plane stress let us denote compression 
by the sign +, and tension by the sign 

Let us also call that state of stress 
which is defined by equal principal 
stresses of the same sign a Jluid stress. 
A material fluid can actually sustain 
only a ■}- fluid stress, but it is convenient 
to include both compression and tension 
under one head as fluid stress, the proper- 
ties of which we shall soon discuss. 

Let us call a state of stress which is 
defined by unequal principal stresses of 
the same sign cm oblique stress. This 
may be taken to include fluid stress as 
the particular case in which the ine- 
quality is infinitesimal. In this state of 
stress there is no plane on which the 
stress is a shear only, and the normal 
component of the stress on any plane 
whatever has the same sign as that of the 
principal stresses. 

Furthermore let us call that state 



108 



TPLE THEORY OF INTERNAL STRESS 



of Stress which is defined by a pair 
of shearing stresses of equal intensity 
and different sign on two planes at 
right angles to each other a right 
shearing stress. We shall have occasion 
immediately to discuss the properties of 
this kind of stress, but we may advan- 
tageously notice one of its properties in 
this connection. It has been seen pre 
viously from considerations of symmetry 
that the principal stresses and planes 
which may be used to define this state 
of stress, bisect the angles between the 
planes of equal shear. Hence in right 
shearing stress the principal stresses 
make angles of 45° with the planes of 
equal shear. We can advance one step 
further by considering the symmetrical 
pt)sition of the planes of equal shear with 
respect to the principal stresses and 
show that the principal stresses in a state 
of right shearing stress are equal but of 
opposite sign. 

We wish to call particular attention 
to fluid stress and to right shearing stress, 
as with them our subsequent discussions 
are to be chiefly concerned : they are the 
special cases in which the principal 
stresses are of equal intensities, in one 
case of the same sign, in the other case 
of different sign. 

Let us call a state of stress which 
is defined by a pair of equal shearing 
stresses of opposite sign on planes 
not at right angles an oblique shear- 
ing stress. The principal stresses, which 
in this case are of unequal intensity 
and bisect the angles between the 
planes of equal shear, are of opposite 
sign. A right shearing stress may be 
taken as the particular case of oblique 
shearing in which the obliquity is in- 
finitesimal. 

We may denote a state of stress as + 
or — according to the sign of its larger 
principal stress. 



be yy two planes at right angles, on 
which the stress at o is normal, of equal 
intensity and of the same sign; then the 
stress on any plane, as x'x\ traversing o 
is normal, of the same intensity and 
same sign as that on xx or yy. 

For co^isider a prism a unit long and 
of infinitesimal cross section having the 
face mn \\ x'x\ then the forces /a; and/y „ 
acting on the faces oni and on are such 
that 

fx '■ fy - '- om : on. 

Now nm^=^ \^ om^ -\- on"^ ^ and the result- 
ant force which the prism exerts against 
nm is 

f^'^/Jx'+fy'^ .-./,:/:: om : mn. 

Bnt fx-^07n is the intensity of the 
stress on xx and f^mn is the intensity 
of the stress on x'x', and these are equal. 
Also by similarity of triangles the result- 
ant/^ is perpendicular to mn. 



r 


Fig. 5 


2/ 


/ 

T 

/ 


7 


f 






/o 


\ 


\ 


•'y 




X 


/ 


\^ 


in 


X 




V 


y 




\ 


f 






Fljid Stress. 



-In Fig. 5 let xx and 



Eight Shearing Stress. — In Fig. 6, 
let XX and yy be two planes at right 
angles to each other, on which the stress 
is normal, of equal intensity, but of 
opposite sign; then the stress on any 
plane, as x'x% traversing o is of the same 
intefisity as that on xx and yy, but its 
obliquity is such that xx and yy respect- 
ively, bisect the angles between the 
direction rr of the resultant stress, and 
the normal y'y' to its plane of action. 

For, if the intensity of the stress on 
x'x' be computed in the same manner as 
in Fig. 5, the intensity is found to be the 
same as that on xx or yy; for the stresses 
to be combined are at right angles and 
are both of. the same magnitude. The 
only difference between this case and 
that in Fig. 5 is this, that one of the 



IN GRAPHICAL STATICS. 



109 



component stresses, that one normal to 
yy say, has its sign the opposite of that 
in Fig. 5. In Fig. 5 the stress on x'x' 
was in the direction y'y' ^ making a cer- 
tain angle yoy' with yy. In Fig. 6 the 
resultant stress on x'x' must then make 
an equal negative angle with yy, so that 
yor^^yoy' . Hence the statement which 
lias been made respecting right shearing 
stress is seen to be thus established. 

Combination and Separation. — Any 
states of stress which coexist at the same 
point and have their principal stresses in 
the same directions xx and yy combine 
to form a single state of stress whose 
principal stresses are the sums of the re- 
spective principal stresses lying in the 
same directions xx and yy : and con- 
versely any state of stress can be separ- 
ated into several coexistent stresses by 
separating each of its two principal 
stresses into the same number of 
parts in any manner, and then grouping 
these parts as pairs of principal stresses 
in any manner whatever. 

The truth of this statement is nec- 
essarily involved in the fact that stresses 
are forces distributed over areas, and that 
as a state of stress is only the grouping 
together of two necessarily related 
stresses, they must then necessarily fol- 
low the laws of the composition and 
resolution of forces. 

For the sake of brevity, we shall use 
the following nomenclature of Avhich the 
meaning will appear without further ex- 
planation. 



The terms applied to 
forces and stresses are : 

Compound, 

Composition, 

Component, 

Resolve, 

Resolution, 

Resultant. 



The terms applied to 
states of stress are : 

Combine, 

Combination, 

Component state, 

Separate, 

Separation, 

Resultant state. 



Other states of stress can be combined 
besides those whose principal stresses 
coincide in direction, but the law of 
combination is less simple than that of 
the composition of forces; such combi- 
nations will be treated subsequently. 



Component Stresses. — Any possible 
state of stress defined by principal 
stresses whose intensities are px and 
Py on the planes xx and yy respect- 
ively is equivalent ' to a combination 
of the fluid stress whose intensity is 
4- ^ ( Pa; + ^9^/ ) on each of the planes xx 
and yy respectively, and the right shear- 
ing stress whose intensity \^-V\^Px— Py) 
on XX and —\{px —/>?/) on yy- 

For as has been shown, the resultant 
stress due to combining the fluid stress 
with the right shearing stress is found 
by compounding their principal stresses. 
Now the stress on xx is 

\{P^^P )-^h{fx-Py)=Px 
and that on yy is 

i(Px -^Py )-\{Px -Py )=Py 

and hence these systems of principal 
stresses are mutually equivalent 

In case py = 0, the stress is complete- 
ly defined by the single principal stress 
Px , which is a simple normal compression 
or tension on xx. Such a stress has been 
called a simple stress. 

A fluid stress and a right shearing 
stress which have equal intensities com- 
bine to form a simple stress. 

It is seen that the definition of a 
state of stress by its principal stresses, 
is a definition of it as a combination of 
two simple stresses which are perpendicu- 
lar to each other. 

There are many other ways in which 
any state of stress can be separated into 
component stresses, though the separa- 
tion into a fluid stress and a right shear- 
ing stress has thus far proved more use- 
ful than any other, hence most of our 
graphical treatment will depend upon it. 
It may be noticed as an instance of a 
different separation, that it was shown 
that the tangential components of the 
stresses on any pair of planes xx and yy 
at right angles to each other are of equal 
intensity but opposite sign. These 
tangential components, then, together 
form a right shearing stress whose prin- 
cipal planes and stresses x'x' and y'y' 
bisect the angles between xx and yy, 
while the normal components together 
define a state of stress whose principal 
stresses are, in general, of unequal in- 
tensity. 



110 



THE THEORY OF INTERNAL STRESS 



Hence any state of stress can be sepa- 
rated into component stresses one of 
which is a right shearing stress on any 
two planes at right angles and a stress 
having those planes for its principal 
planes. 

The fact of the existence of conjugate 
stresses points to still another kind of 
separation into component stresses. 



PROBLEMS IN PLANE STRESS. 

Problem 1. — When a state of stress is 
defined by principal stresses which are 
of unequal intensity and like sign, i.e., in 
a state of oblique stress, to find the in- 
tensity and obliquity of the stress at o 
on any assumed plane in the direction 
uv. 



Fig. 7. 




In Fig. Y let the principal stresses at o 
be a on yy and h on xx ; and on some 
convenient scale of intensities let oa=^a 
and oJ) = h. Let uv show the direction 
of the plane through o on which we are 
to find the stress, and make on perpendic- 
ular uv. Make oa'^^^oa and oh'^^oh. 
Bisect a'h' at m, then on^=^\(a ■\-})) and 
7ia'^=\{a—b). Make xol:=xon and com- 
plete the paralellogram nomr\ then is 
the diagonal or^r the resultant stress 
on the given plane in direction and in- 
tensity. 

The point r can also be obtained more 
simply by drawing b'r\\xx and a'r \\ yy. 

We now proceed to show the correct- 
ness of the constructions given and to 
discuss several interesting geometrical 
properties of the figure which give to it 
a somewhat complicated appearance, 
which complexity is, however, quite un- 
necessary in actual construction, as will 
be seen hereafter. It has been shown 



that a state of stress defined by its two 
principal stresses a and b can be separ- 
ated into a fluid stress having a normal 
intensity ^ (a -1-5) on every plane, and a 
right shearing stress whose principal 
stresses are 4-J(a— Z>) and —^{a—b) re- 
spectively. 

Since the fluid stress causes a normal 
stress on any given plane, its intensity is 
rightly represented by on=-^(a + b), 
which is the amount of force distributed 
over one unit of the given plane. Since, 
further, it was shown that a right shear- 
ing stress causes on any plane a stress 
with an obliquity such that the principal 
stress bisects the angle between its direc- 
tion and the normal to the plane, and 
causes a stress of the same intensity on 
every plane, we see that om=^{a—b) 
represents, in direction and amount, the 
force distributed over one unit of the 
given plane which is due to the right 
shearing stress. 



IN GRAPHICAL STATICS. 



Ill 



To find the resultant stress we have 
only to compound the forces ooi and om, 
which give the resultant or:=r. 

The obliquity nor is always toward 
the greater principal stress, which is here 
assumed to be a. 

It is seen that in finding r by this 
method it is convenient to describe one 
circle about o with a radius o/=^(a + 6) 
and another with a radius og--^{a—b), 
after which any parallelogram tnn can 
be readily completed. Let ?^r and mr 
intersect xx and yy in hk and ij respect- 
ively; then we have the equations of 
angles, 

noh^?iho^=^Jcno, nok=9iko=^h7io, 
'moi=7nio=^^jmOj nioj=mJo:=^i7nOy 
hence hn=kn=^on=:-^(a + b) 

and rk=r/=a, rh=ri=b. 

It is well known that a fixed point r 
on a line of constant length as hk=a-\-by 
or ij=a—b describes an ellipse, and 
such an arrangement is called a trammel. 
If X and y are the coordinates of the 
point r, it is evident from the figure that 
x=acosx7i, y=b sin xn, in which xn 
signifies the angle between xx and the 
normal on, 

x^ y^ 
• *. -^ + T^=1 is the equation of the stress 
a b 

ellipse which is the locus of r; and xn is 

then the eccentric angle of r. Also, since 

noh=:nho, nb'r=.nrb' \ hence b'r \\ icccand 

a'r II yy determine r. 

In this method of finding r it is con- 
venient to describe circles about o with 
radii a and 6, and from a' and b' where 
the normal of the given plane intersects 
them find r. 

We shall continue to use the notation 
employed in this problem, so far as ap- 
plicable, so that future constructions 
may be readily compared with this. It 
will be convenient to speak of the angle 
xon as xn^ nor as nr^ etc. 

Pkoblem 2. — When a state of stress is 
defined by principal stresses of unequal 
intensity and unlike sign, i.e. in a state 
of oblique shearing stress, to find the in- 
tensity and obliquity of the stress at o 
on any assumed plane having the direc- 
tion uv. 



In Fig. 8 the construction is effected 
according to both the methods detailed 
in Problem 1, and it will be at once ap- 
prehended from the identity of notation. 

Since a and b are of unlike signs a-\-b 
=zon is numerically less than a—b^^a'b'. 

The results of these two problems are 
expressed algebraically thus: 

T^=l{a^bY-V\{a-bY^\{a^-b'')Q,Qs^xn 

,'. r'=^[a' + b' + (a'-b')G0s2xnJ 

or, r^ — a^ cos'^xn + b^ sin^xn. 

Fig. 8. 




If r be resolved into its normal and 
tangential components oi=n and rt=:t 

then, n=i[a -^-b-h {a— b) cos 2xn\, 

or, n=^a c.os^xn-\-b sm^xn, 
and, 

t^:^^{a—b)sm 2xn=i{a—b)sm xn Qosxn, 

It is evident from the value of the 
normal component n, that the sum of the 
normal components on any two i)lanes at 
right angles to each other is the same 
and its amount is a + b: this is also a 
general property of stress in addition to 
those previously enumerated. 

., t a—b 

Also tannr-— -= ^ 

n a cot xn-Vb tan xn 

The obliquity nr can also be found 
from the proportion 

sin nr : ^{a—b) : : sin 2xn : r. 

In the case of fluid stress the equations 
reduce to the more simple forms: 

a=^b=r=znj t=:0 
For right shearing stress they are: 
a=—b=-\-r, n= :ta cos rn, 
t=:t a sin rn, rn=2 xn. 



112 



THE THEOEjr OF INTERNAL STRESS 



And for simple stress they become: 
5=0, r=« cos rUy n=a cos'ni, 
t=a sin rn cos ni, rn=xn. 

Problem 3. — In any state of stress 
defined by its principal stresses, a and b, 
to find the obliquity and plane of action 
of the stress having a given intensity r 
intermediate between the intensities of 
the principal stresses. 

To find the obliquity nr and the direc- 
tion tiv let Fig. 7 or 8 be constructed as 
follows: assume the direction uv and its 
normal oUj and proceed to determine the 
position of the principal axes with re- 
spect to it. Lay off oa'=a, ob' = b, in 
the same direction if the intensities are 
of like sign, in opposite directions if un- 
like. Bisect a'b' at n, and on a'b^ as a 
diameter draw the circle a^rb'. Also, 
about o as a center and with a radius 
07'=r draw a circle intersecting that pre- 
viously drawn at r; then is nr the re- 
quired obliquity; and xxW^r, yy\\a'r 
are the directions of the principal stresses 
Avith respect to the normal on. 

Problem 4. — In a state of stress de- 
fined by two given obliquities and in- 
tensities, to find the principal stresses, 
and the relative position of their planes 
of action to each other and to the 
principal stresses. 

Fig. 9. 




In Fig. 9 let wr^, nr, be the given 
obliquities measured from the same nor- 



mal on^ and c>rj=r„ or^-=^r^ the given in- 
tensities. As represented in the figure 
these intensities are of the same sign, but 
should they have different signs, it will 
be necessary to measure one of them 
from o in the opposite direction, for a 
change of sign is equivalent to increas- 
ing the obliquity by 180°, as was pre- 
viously shown. 

Join r/g and bisect it by a perpendicu- 
lar which intersects the common nor- 
mal at n. About n describe a circle 
r^r^a'b'\ then oa' = a, ob'=b^ a'r^^ 6V,, 
are the directions of the principal stresses 
with respect to r^ and b'r^^ a'r^ with re- 
spect to r-j, ^.e., obW^-=-xn^ and ob'r^-=xn^ 

.*. n^n^^=^ob'r^—ob'r^^=rJb'r^:=^r^a'r^ 

In case the given obliquities are of op- 
posite sign, as they must be in conjugate 
stresses, for example, it is of no conse- 
quence, in so far as obtaining principal 
stresses a and b is concerned, whether 
these given obliquities are constructed on 
the same side of o?^, or on opposite sides 
of it; for a point on the opposite side of 
on, as r^\ and symmetrically situated with 
respect to r^, must lie on the same circle 
about n. But in case opposite obliquities 
are on the same side of on we have 
n^n^-=ob'r^ ■\- ob'r^=^rJ)'r^', 

It is unnecessary to enter into the 
proof of the preceding construction as 
its correctness is sufliciently evident from 
preceding problems. 

Thq algebraic relationships may be 
written as follows: 

l(a-hy=::l[a-^by + r^'-r^(a-\-b)Qo^n/^ 

i (a — Z>) ' = 4 (a -I- ^>) -I- r^' — r' (a -f ^>) cos w,r, 

.*. (a + ^)(^iCOS n/j— r^cos 'i\f^=i''^—r* 

Also {a—b)Q0^2xn^'\-a-\-b^=2r^QO^n^r^ 

{a—b)cos 2xn^ + a-\-b=2r^coan^r^ 

which last equations express twice the 
respective normal components, and from 
them the values of xn^ and xn^ can be 
computed. 

Problem 5. — If the state of stress be 
defined by giving the intensity and 
obliquity of the stress on one plane, and 
its inclination to the principal stresses, 
and also the intensity of the stress on a 
second plane and its inclination to the 
principal stresses, to find the obliquity of 



IN GRAPHICAL STATICS. 



113 



the stress on the second plane, and the 
magnitude of the principal stresses. 

Let the construction in Fig. 9 be 
effected thus: from the common normal 
on lay off or^ to represent the obliquity 
and intensity of the stress on the first 
plane; draw od so that nod=xn^—xn^ 
the difference of the given inclinations 
of the normals of the two planes; 
through r^ draw^j^^ perpendicular to od; 
about as a center describe a circle with 
radius 7\ the given intensity on the 
second plane, and let it intersect r^7\ at 
r^ or r/, then is 9ir^ the required obliquity. 
This is evident, because 

, \ nod:= one = \ {om\ + onr^ 

= 1 80' — {xn^—xn^ 

If xn^ and xn^ are of different sign 
care must be taken to take their alge- 
braic sum. 

The construction is completed as in 
Problem 4. 

Problem 6. — In a state of stress de- 
fined by two given obliquities and either 
both of the normal components or both 
of the tangential components of the in- 
tensities, to find the principal stresses 
and the relative position of the two 
planes of action. 

If in Fig. 9 the obliquities nr^^ nr^^ and 
the normal components ot^-=n^^ ot^-=^n^ 
are given, draw perpendiculars at t^ and 
t^ intersecting or^ and or^ at r, and r^ re- 
spectively. 

If the tangential components i^r^-=^t^ 
and^^rj=^2 ^^® given instead of the nor- 
mal components, draw at these distances 
parallels to on which intersect oi\ oi\ at 
r^i\ respectively. Complete the con- 
struction in the same manner as before. 

Problem 7. — In a state of stress de- 
fined by its principal stresses a and 5, to 
find the positions and obliquities of the 
stresses on two planes at right angles to 
each other whose stresses have a given 
tangential components. 

Fig. 9, slightly changed, will admit of 
the required construction as follows: lay 
off on the same normal on, oa'=^a, oh'^h\ 
bisect a'h' at n ; erect a perpendicular 
ne-=t to a'h' at n ; draw through e a 
parallel r^r^ to on intersecting oi\ and 



oi\ at r^ and r^ respectively. Then the 
stresses or^^=^r^, or^=^r^ have equal tan- 
gential components, and as previously 
shown these belong to planes at right 
angles to each other provided these tan- 
gential components are of opposite sign. 
So that when we find the position of the 
planes of action, one obliquity, as nr^j 
must be taken on the other side of oti, 
as ni\\ The rest of the construction is 
the same as that already given. , 

Problem 8. — In a state of stress de- 
fined by its principal stresses, to find the 
intensities, obliquities and planes of 
action of the stresses which have maxi- 
mum tangential components. 

In Fig. 9 make o«'=a, oh'^=^h and 
describe a circle on a'h' as a diameter; 
then the maximum tangential component 
is evidently found by drawing a tangent 
at r parallel to on^ in which case t^a—h, 
and rh\ ra the directions of the 
principal stresses make angles of 45° 
with on^ which may be otherwise stated 
by saying that the planes of maximum 
tangential stress bisect the angles be- 
tween the principal stresses; or con- 
versely the principal stresses bisect the 
angles between the pair of planes at 
right angles to each other on which the 
tangential stress is a maximum. 

It is unnecessary to extend further the 
list of problems involving the relations 
just employed as they will be readily 
solved by the reader. 

In particular, a given tangential and 
normal component may replace a given 
intensity and obliquity on any plane. 

We shall now give a few problems 
which exhibit specially the distinction 
between states of stress defined by 
pjincipal stresses of like sign and by 
principal stresses of unlike sign, {i.e. the 
distinction between oblique stress and 
oblique shearing stress). 

Problem 9. — In a state of stress de- 
fined by like principal stresses, to find 
the inclination of the planes on which 
the obliquity of the stress is a maximum, 
to find this maximum obliquity and the 
intensity. 

In Fig. 10 let oa'=a, bh' = h, the 
principal stresses; on a'h' as a diameter 
describe a circle; to it draw the tangent 
oi\\ then m\ is the required maximum 



114 



THE THEORY OF INTERNAL STRESS 



obliquity and or^ the required intensity. 
It is evident from inspection that in the 
given state of stress there can be no 
greater obliquity than nr^. The direc- 
tions of the principal axes are h'r^^ a'r^ 
as has been before shown. 

There are two planes of maximum 
obliquity, and or^' represents the second; 
they are situated symmetrically about 
the principal axes. 

♦ Bisect nr^ by the line od, then 

oa'r^-=yn .*. oni\=2yn, but 

owr„ + woro=90° or, 2yn + nr^=Q0° 

.'. \nr^-\yn^=^b°, but 

odi\=.doa' -\-oa'd ,\ odr^--4:5''y 

hence the line bisecting the angle of 
maximum obliquity bisects also the 
angle between the principal axes. This 
is the best test for the correctness of the 
final position of the planes of maximum 
obliquity with reference to the principal 
axes. 

Fig. 10. 




Problem 10. — In a state of stress de- 
fined by its maximum obliquity and the 
intensity at that obliquity, to find the 
principal stresses. 

In Fig. 10 measure the obliquity nr^ 
from the normal on and at the extremity 
of or^-=r^ erect a perpendicular inter- 
secting the normal at n. Then complete 
the figure as before. The principal 
axes make angles of 45° at o with od 
which bisects the obliquity nr^. 

The algebraic statement of Problems 
9 and 10 is: 



sm nr^ 



a—h 

a + b 



= — cos 2xnj r*=:ab. 



r^=a cot xn=b tan xn, .*. a=b tan*xn 

The normal and tangential compo- 
nents are: 



2r' 



"""-^T^' 



^ r^(a-b) 
a-hb 



Problem 11. — When the state of 
stress is defined by like principal stresses, 
to find the planes of action and intensi- 
ties of a pair of conjugate stresses having 
a given common obliquity less than the 
maximum. * 

In Fig. 10 let nr=nr^ be the given 
obliquity; describe a circle on a'b' as a 
diameter; then or^-=r^^ or^=r^ are the 
required intensities. The lines aV^, b'r^ 
show the directions of the principal axes 
with respect to or^, and ar\ b'r\ with 
respect to or^=or^. The obliquities of 
conjugate stresses are of opposite sign, 
and for that reason r^' is employed for 
finding the position of the principal 
stresses. The algebraic expression of 
these results can be obtained at once 
from those in Problem 4. 

Problem 12, — When the state of stress 
is defined by the intensities and common 
obliquity of a pair of like conjugate 
stresses, to find the principal stresses and 
maximum obliquity. 

This is the case of Problem 4, so far as 
finding the principal stresses is concerned, 
and the maximum obliquity is then found 
by Problem 9. The construction is given 
in Fig. 10. 

Problem 1*^.— Let the maximum ob- 
liquity of a state of oblique stress be 
given, to find the ratio of the intensities 
of the pair of conjugate stresses having 
a given obliquity less than the maxi- 
mum. 

In Fig. 10 let m\ be the given maxi- 
mum obliquity, and 7i r^ the given ob- 
liquity of the conjugate stresses. At 
any convenient point on or^^ as r^ erect 
the perpendicular r^n, and about n (its 
point ot intersection with on) as a center 
describe a circle with a radius m\ which 



IN GRAPHICAL STATICS. 



115 



outs wr^ at r, and r^; then or-^oi\:=r^ 
-f-^2 is the required ratio. 

It must be noticed that the scale on 
which oi\ and 07\ are measured is un- 
known, for the magnitude of the princi- 
pal stresses is unknown although their 
ratio is oh' -^oa\ In ordet to express 
these results in formulae, let r represent 
either of the conjugate stresses, then as 
previously seen 

\{a—hY=\ {a-^hy^r'^—r{ct-\-h) cos nr 

,'. 2r=^(«-f-^)cos nr±. 

[(a-f 5)''cos'wr — 4a6]>^ 

Call the two values of r, r^ and t^\ 
and as previously shown T^^=-r^r^\ also 

COS. nr=r^-^\{a^h) 
r^_Q.o?>nr — {Gos^nr—Gos^7i9\)y^ 

7*2 COS nr + {C0ii'^7l7' — G0S^nr^)>^ 

When 917^=0 the ratio becomes 
b 1 — sin nr„ 



a 



1 -I- sin nr„ 



Pkoblem 14. — In a state of stress 
defined by unlike principal stresses, to 
find the inclination of the planes on 
which the stress is a shear only, and to 
find its intensity. 

In Fig. 11 let oa'=«, ob^ = b, the 
given principal stresses of unlike sign; 
on a'b' as a diameter describe a circle; 
at o erect the perpendicular or^ cutting 
the circle at 7\i, then is or^=^?\ the re- 
quired intensity, and b'r^, a'7\ are the di- 
rections of the principal stresses. 

It is evident from inspection that there 
is no other position of r^ except r/ 
which will cause the stress to reduce to 
a shear alone. Hence as previously 
stated the principal stresses bisect the 
angles between the planes of shear. 

Problem 15.— In a state of stress de- 
fined by the position of its planes of 
shear and the common intensity of the 
stress on these planes, to find the princi- 
pal stresses. 

In Fig. 11 let or^—r^, the common in- 
tensity of the shear, and orJb'^=.xn^ 
or^a'=^yn the given inclinations of a 
'plane of shear; then oa'^=a and ob' = b^ 
the principal stresses. 

The algebraic statement of Problems 



14 and 15, when n^ denotes the normal 
to a plane of shear, is: 

Fig. 11. 




a-\-b 



a—b 



= — cos 2xn^ 



T^=—ab—t, 



r^ = ^a cotxn^=4-btSin xn^^a=-bX,2^n^x7i^ 

Problem 16.— When the state of 
stress is defined by unlike principal 
stresses, to find the planes of action and 
intensities of a pair of conjugate stresses 
having any given obliquity. 

In Fig. 11 let nr^ be the common ob- 
liquity, oa'=^a, ob'=^b, the given princi- 
pal stresses. On ab\ as a diameter, 
describe a circle cutting 07\ at r^ and r 



2> 



then or^=r^, or^=r^ are the required in- 
tensities. Also, since the obliquities of 
conjugate stresses are of unlike sign, the 
lines r/a% r/5^ show the directions of the 
principal stresses with respect to on^y 
and r^a', 7\b' with respect to on^. 

Problem 11. — When the state of stress 
is defined by the intensities and common 
obliquities of unlike conjugate stresses, 
to find the principal stresses and planes 
of shear. 

In finding the principal stresses this 
problem is constructed as a case of 
Problem 4, and then the planes of shear 
are found by Problem 14. The con- 
struction is given in Fig. 11. 

Problem 18.— Let the position of the 



116 



THE THEORY OF INTERNAL STRESS 



planes of shear be given in a state of 
oblique shearing stress, to find the ratio 
of the intensities of a pair of conjugate 
stresses having any given obliquity. 

In Fig. 11 at any convenient point r^ 
make orJ)'^^xn^ or^a'=^yn^ the given 
angles which fix the position of the 
planes of shear. On a'h' as a diameter 
describe a circle; make nr^ equal to the 
common obliquity of the conjugate 
stresses; then is or^-r-or^^=^r^^r^\hQ ratio 
required. 

The ratio may be expressed as in 
Problem 13, and after reducing by the 
relations 

r^=—ah^ r„-j-J(« + 5) = — tan2a!^, 
we have, 

r, cos nr + (cos'^^r^-tan''2£c?^J^ 
r^ ~ cos nr — (cos^?2rH-tan'2a;w.„)>^ 

When nr=0 the ratio becomes 
a_H-cos 2x71^ 
^~1— cos 2xn^ 

COMBINATION AND SEPARATION OF STATES 
OF STRESS. 

Problem 19.— When two given states 
of right shearing stress act at the same 
point, and their principal stresses have a 
given inclination to each other, to com- 
bine these states of stress and find the 
resultant state. 

In Fig. 12 let ox^, ox^ denote the di- 
rections of the two given principal + 
stresses, and let a^=^on^, a^-=on^ repre- 




sent the position and magnitude of these 
principal stresses. Since the given 
stresses are right shearing stresses 
<^j = — 5j, ^2=— ^2 and the respective 
planes of shear bisect the angles between 
the principal stresses. Now it has been 
previously shown that the intensity of 
the stress caused by the principal stresses 
<Xj = — 6j is the same on every plane 
traversing o: the same is true of the 
principal stresses a^-= — h^ : hence, when 
combined, they together produce a stress 
of the same intensity on every plane 
traversing o. This resultant state of 
stress evidently .does not cause a normal 
stress on every plane, hence the result- 
ant state must be a right shearing stress. 

Let us find its intensity as follows : 
The principal stresses a^=—b^ cause a 
stress on^ on the plane .Vi^/i* ^"^ ^^^ princi- 
pal stresses a^=—b^ cause a stress om^ on 
the same plane in such a direction that 
x^om^=x^ox^, as has been before shown. 
Complete the parallelogram n^om^r^; 
then oi\ represents the intensity and di- 
rection of the stress on y^y^. But the 
principal stresses bisect the angles be- 
tween the normal and the resultant in- 
tensity, therefore, oa?, which bisects 
x^oi\^ is the direction of a principal stress 
of the resultant state, and orz=or^=a is 
the intensity of the resultant stress on 
any plane through o. 

The same result is obtained by finding 
the stress the plane y^y^y in which case 
we have on^-=a^ acting normal to the 
plane, and om^^^a^ in such a direction 
that x^o')n^.^=iX^ox^. The sides and angles 
of n^om^r^ and n^om^i\ are evidently 
equal, hence the resultants are the same,^ 
or^:=or^=:a, and ox bisects x^or^. 

The algebraic solution of the problem^ 
is expressed by the equation, 

a^=a^ + a^ -{-2a^a^ cos 2 x^x,^, 

from which a may be found, and, finally^ 
the position of or is found from the pro" 
portion, 

sin 2xx^ : «2 • ► ^i" ^aja;^ : a^',', sin 2x^x^ : a. 

Problem 20. — When any two states 
of stress, defined by their principal 
stresses, act at the same point, and their 
principal stresses have a given inclina- 
tion to each other, to combine these* 
states and find the resultant state. 

Let a^, 6j, and a^, b^ be the given prin- 



IN GRAPHICAL STATICS. 



117 



cipal stresses, of which a. and a^ have 
the same sign and are inclined at a 
known angle x^x^,, but in so taking a^ 
and a^ they may not both be numerically 
greater than b^ and h^ respectively. 

Separate the pair of principal stresses 
a^b^ into the fluid stress -\-\(a^ + b^, and 
the right shearing stress ±.^{a^ — b^) as 
has been previously done; and in a simi- 
lar manner the principal stresses a^ b^ 

into +1(^2 + ^2) ^."^ d-i(<^2— ^J- Then 
the combined fluid stresses produce a 

fluid stress of +^(«i + ^i + «2 + ^2) ^^ 
every plane through 0; and the com- 
bined right shearing stresses cause a 
stress whose intensity and position can 
be found by Problem 19. . 

The total stress is obtained by com- 
bining the total fluid stress w;,ith the re- 
sultant right shearing stress. 

Of course, any greater number of 
states of stress than two, can be com- 
bined by this problem by combining the 
resultant of two states with a third state 
and so on. 

The algebraic expression of the com- 
bination of any two states of stress is as 
follows : 

[a + b) = {a^ + b, + a^ + b;), 

(a-by={a-by + (a-b:)^ 

+ 2{a-b^) {a^-bj cos 2x^x^, 

.'. a=i{a^-]-b^-\-a^ + b^ + [{a-by 

+ K-^2)" + 2(a -5 J («,-^)cos 2x^x^]y^), 

b=^(a^ + b^-i-a^ + b,-[{a-b;)^ + (a-b:)^ 
+ 2{a-b^)(a^-b^)Gos 2x^x^]y^), 

in which a and b are the resultant prin- 
cipal stresses. Also, sin 2xx^: a^—b„ 

: : sin 2xx^-. ci^ — b^ : : sin 2x^x^: a—b. 

Pkoblkm 21. — In a state of stress 
defined by the stresses upon two planes 
at right angles to each other, to find the 
principal stresses. 

Let the given stresses be resolved into 
tangential and normal components; it 
has been shown that the tangential com- 
ponents upon these planes are of equal 
intensity and unlike sign. Let the in- 
tensity of the tangential component be 
at, and that of the normal components 
an and bn respectively. The tangential 
components together constitute a state 
of right shearing stress of which the 
given planes are the planes of shear, 



and the principal stresses bisect the 
angles between the given planes. 

Separate the remaining state of stress 
into the fluid stress -\-^{an + bn) and 
the right shearing stress ±\(an — bn)y 
and combine this last right shearing 
stress with that due to the tangential 
components. The final result is found, 
just as in Problem 20, by combining the 
fluid stress |(«n + bn) with the resulting 
right shearing stress. 

This problem can also be solved in a 
manner similar to that employed in 
Problem 6. 

The result is expressed by the equa- 
tions, 

a + b=an + bn, 

{a—by=i(an — bny + 4:at ' 

for the angle which has been heretofore 
denoted by x^x^ is in this case 45° .'. cos 
2i:CjCC2 = 

.-. ci=^{an -\-bn + [(an -bn)' + 4af=']>^J 
b=i{an + bn -[{an - bnY-h^atY') 

sin. 2xx^ : 2a t : : sin. 2xx^ : an — bn 

: : 1 : a — b^ 
but 2a!a;j = 90° — 2a!iC2 , 

.*. tan 2xx^ = 2at -^ {an — bn)> 

Problem 22. — In a state of stress 

deflned by two simple stresses which act 

at the same point and have a given 

inclination to each other, to combine 

them and find the resultant state. 

It has been previously mentioned that 
any simple stress as a^ can be separated 
into the fluid stress +i«i and the right 
shearing stress ±J«i, as it is simply a 
case *in which b^=^0. Hence the simple 
stresses a^, a^ can be combined as a spe- 
cial case of Problem 20, in which b^ and 
b^ vanish. The results are expressed 
algebraically as follows: 

a^b=a^ + a^^ 
{a—b)^=a^^-\-a^^-\-2a^a^ cos 2x^x^ 

,'. ab=ia^a^{l — cos 2x^x^) 
.'. ab^^a^a^ sin^x^x^. 

Since a simple compression or tension 
produces a simple stress in material, this 
problem is one of frequent occurrence, 
for it treats the superposition of two, 
and hence of any number of simple 
stresses lying in the same plane. 

This problem is of such importance 
that we think it useful to call attention 



118 



THE THEORY OF INTERNAL STRESS 



to another solution of it, suggested by 
the algebraic expressions just found. 
In Fig. 13 let 

o'a'^^j, o'h'=^a^ .*. o'r' = ^a^a^=^oi. 

Now, if oir=x^x^^ then or=.o'r' sin x^x^ 

.-. or'^:=^oa'.ol>^-=o'a'.o'b' ^m^x^x^ 

.'. oa'=a and oh' =.h. 




This solution is treated more fully in 
Problem 23. 

Problem 23. — When a state of stress 
is defined by its principal stresses, it is 
required to separate it into two simple 
stresses having a given inclination to 
each other. 

It was shown in Problem 22 that 
a + d=:a^4-a,, and ab=a^a^^\xi x^x^. 

Let us apply these equations in Fig. 
13 to effect the required construction. 
Make oa' = a^ oh'=h\ then a'h'=a^ + a^. 
At erect a perpendicular to a'h' cut- 
ting the circle of which a'b' is the dia- 
meter at r; then or^ = o,h, the product of 
the principal stresses. Also make a'oi 
•=x^x^ the given inclination of the sim- 
ple stresses, and let ri \\ a'h' intersect oi 
at ^ ; then or=ioi sin x^x^ .*. oi"^ = a^a^. 

Make oj—oi and draw jr' \\ a'h' ^ then 

o'r' = oi, and o'a',o'h'=zo'r'^, 
•.• o'a'-=a^ and o'h'=za^^ 

the required simple stresses. This con- 
struction applies equally whether the 
given principal stresses are of like or 
unlike sign, and also equally whether 
the two simple stresses are required to 
have like or unlike signs. 

Problem 24. — When a state of stress 
is defined by its principal stresses, to 
find the inclination of two given simple 
stresses into which it can be separated. 



In Fig, 13 let oa' = a, oh'^h be the 
intensities of the principal stresses, and 
o'a'^=a^, o'b'=^a^ be the intensities of the 
given simple stresses. It has been 
already shown that a-\-h^=a^-\-a^. Draw 
the two perpendiculars or and o'r'\ 
through ?' draw ri \\ a'h' \ make oi=oj 
= o'r'i then is oir=ioa' the required 
inclination, for it is such that 

ah=a^a^ sin^'a^^ir^ 

Problem 25. — To separate a state of 
right shearing stress of given intensity 
into two component states of right shear- 
ing stress whose intensities are given, and 
to find the mutual inclination of the 
principal stresses of the component 
states. 

In Fig. l2, about the center o, describe 
circles with radii 09i^ — a^^ on^=a^y the 
given component intensities; and also 
about o at a distance or^ = a, the given 
intensitv. Also describe circles with radii 
r^7n^=^on^, r^n^=09i^ cutting the first 
mentioned circles at m^ and 7i^: then is 
i9i^07n^=x^x^ the required mutual inclina- 
tion of the principal stresses of the com- 
ponent states. This is evident from 
considerations previously adduced in con- 
nection with this figure. The relative 
position of the principal stresses and 
principal component stresses is also read- 
ily found from the figure. 

Problem 26.— In a state of right 
shearing stress of given intensity to sep- 
arate it into two component states of 
right shearing stress, when the intensity 
of one of these components is given and 
also the mutual inclination of the princi- 
pal stresses of the component states. 

In Fig. 12, about the center o describe 
a circle rr with radius or=a, the inten- 
sity of the given right shearing stress, 
and at 7i^, at a distance on^=a^ from o 
which is the intensity of the given com- 
ponent, make x^)i^r^=2x^x^f twice the 
given mutual inclination ; then is n^r^ 
the distance from n^ to the circle rr the 
intensity of the required component 
stress. The figure can be completed as 
was done previously. 

It is evident, when the component a^ 
exceed «, that there is a certain maxi- 
mum value of the double inclination, 
which can be obtained by drawing n^r^ 



IN GRAPHICAL STATICS. 



119 



tangent to the circle rr, and the given in- 
clination is subject to this restriction. 

Other problems concerning the com- 
bination and separation of states of 
stress can be readily solved by methods 
like those already employed, for such 
problems can be made to depend on the 
combination and separation of the fluid 
stresses and right shearing stresses into 
which every state of stress can be sep- 
arated. 

PROPERTIES OF SOLID STRESS. 

We shall call that state of stress at a 
point a solid stress which causes a stress 
on every plane traversing the point. In 
the foregoing discussion of plane stress 
no mention was made of a stress on the 
plane of the paper, to which the plane 
stress was assumed to be parallel. It is, 
evidently, possible to combine a simple 
stress perpendicular to the plane of the 
paper with any of the states of stress 
heretofore treated without changing the 
stress on any plane perpendicular to the 
paper. 

Hence in treating plane stress we have 
already treated those cases of solid stress 
which are produced by a plane stress 
combined with any stress perpendicular 
to its plane, acting on planes also per- 
pendicular to the plane of the paper. 

We now wish to treat solid stress in a 
somewhat more general manner, but as 
most practical cases are included in plane 
stress, and the difficulties in the treat- 
ment of solid stress are much greater 
than those of plane stress, we shall make 
a much less extensive investigation of its 
properties. 

Conjugate Stresses. — Let xx,, yy, zz 
be any three lines through 0/ now, if 
any state of stress whatever exists at 0, 
and XX be the direction of the stress on 
the plane yoz, and yy that on zox, then 
is zz the direction of the stress on xoy : 
i.e., each of these three stresses lies in 
the intersection of the planes of action of 
the other two. 

Reasoning like that employed in con- 
nection with Fig. 1, shows that no other 
direction than that stated could cause 
internal equilibrium; but a state of stress 
is a state of equilibrium, hence follows 
the truth of the above statement. ' 



Tangential Components. — Let xx, 
yy, zz be rectangular axes through 0; 
then, whatever may be the state of stress 
at o, the tangential components along xx 
and yy are equal, as also are those along 
yy and zz, as well as those along zz and 

XX. 

The truth of this statement flows at 
once from the proof given in connection 
with Fig. 3. 

It should be noticed that the total 
shear on any plane xoy, for example, is 
the resultant of the two tangential com- 
ponents which are along xx and yy re- 
spectively. 

State of Stress. — Any state of solid 
stress at is completely defined, so that 
the intensity and direction of the stress 
on any plane traversing o can be com- 
pletely determined, when the stresses on 
any three planes traversing are given 
in magnitude and direction. 

This truth appears by reasoning simi- 
lar to that employed with Fig. 4, for the 
three given planes with the fourth en- 
close a tetrahedron, and the total dis- 
tributed force acting against the fourth 
plane is in equilibrium with the resultant 
of the forces acting on the first three. 

Principal Stresses. — In any state of 
solid stress there is one set of three con- 
jugate stresses at right angles to each 
other, i.e. there are three planes at right 
angles on which the stresses are normal 
only. 

Since the direction of the stress on any 
plane traversing a given point can 
only change gradually, as the plane 
through changes in direction, it is 
evident from the directions of the 
stresses on conjugate planes that there 
must be at least one plane through on 
which the stress is normal to the plane. 
Take that plane as the plane of the 
paper; then, as proved in plane stresses, 
there are two more principal stresses 
lying in the plane of the paper, for the 
stress normal to the plane of the paper 
has no component on any plane also 
perpendicular to the paper. 

Fluid Stress. — Let the stresses on 
three rectangular planes through o be 



120 



THE THEORY OF INTERNAL STRESS 



normal stresses of equal intensity and 
like sign; then the stress on any plane 
through is also normal of the same in- 
tensity and same sign. 

This is seen to be true when we com- 
bine with the stresses already acting in 
Fig. 5, another stress of the same inten- 
sity normal to the plane of the paper. 

Right Shearing Stress.— Let the 
stresses on three rectangular planes 
through be normal stresses of equal 
intensity, but one of them, say the one 
along icic, of sign unlike that of the other 
two; then the stress on any plane through 
0, whose normal is x'x' ^ is of the same 
intensity and lies in the plane xox' in 
such a direction rr that xx and the plane 
yz bisect the angles in the plane xox' be- 
tween rr and its plane of action, and 
vox' respectively. 

The stress parallel to yz is a plane 
fluid stress, and causes therefore a normal 
stress on the plane xox' . Hence the re- 
sultant stress is in the direction stated, 
as was proved in Fig. 6. 

Component States of Stress. — Any 
state of solid stress, defined by its prin- 
cipal stresses abc along the rectanglar 
axes of xyz respectively, is equivalent to 
the combination of three fluid stresses, 
as follows: 

^(a-fJ) alongtc andy, — ^(a + 5) along z\ 
\{c^-a) along z and x^—\(c-\-a) along y\ 
J(6-i-c) along y and 2,— i(^ + c) along y\ 

For these together give rise to the fol- 
lowing combination: 

\(a-\-h)-^\{c^-d)—\{})^c)—a, along cc; 
\(a^-'b) — \{c-\-a)^\{})^c) = h, along y; 
^(a-i-5)+J(c-|-a)-fi(^>-fc)=c, along x. 

In case J=0 and c=0 this is a simple 
stress along x. 

Component Stresses. — Any state of 
solid stress defined by its principal 
stresses can also be separated into a fluid 
stress and three right shearing stresses^ 
as follows: 

i(a -f ?> + c) along x, y, z ; 



\{a—'b — c) along cc, and 

—\{Gi -b—c) along 2/ and z; 

\(b — ('—a) along y, and 

— iip — c—a) along z and x ; 

\{c — a—b) along 2, and 

—\{c — a-—h) along x and y ; 

It will be seen that the total stresses 
along xyz are abc respectively. This 
system of component stresses is remarka- 
ble because it is strictly analagous in its 
geometric relationships to the trammel 
method used in plain stress. We shall 
simply state this relationship without 
proof, as we shall not use its properties 
in our construction. 

If the distances po.^=:.a, pb^-=b, pc^^=c 
be laid off along a straight line from the 
point p, and then this straight be moved 
.so that the points a^ b^ c^ move respec- 
tively in the planes yz, zx, xy ^' then p 
will describe an ellipsoid, as is well 
known, whose principal semiaxes are 
along xyz, and are abc respectively. 
Now the distances pa^^pb^, pc^, maybe 
laid off in the same direction from p or 
in different directions; so that, in all, 
Tour different combinations can be made, 
either of which will describe the same 
ellipsoid. But the position of these 
four generating lines through any as- 
sumed point x^y^z^ of the ellipsoid is such 
that their equations are 

^{x-x;) =. +- {y-y,) = ±-(2-2,) 



X 



■y. 



Now if the fluid stress ^(a-t-6-f c)=or, 
be laid off along the normal to any plane, 
i.e. parallel to that generating line which 
in the above equation has all its signs 
positive, and the other three right shear- 
ing stresses r^r.^, r^r^, r^9\ be laid off 
successively parallel to the other generat- 
ing lines, as was done in plane stresses, 
the line oi\ will be the resultant stress on 
the plane. 

problems in solid stress. 

Problem 27. — In any state of stress 
defined by the stresses on three rectangt- 
lar planes, to find the stress on any given 
plane. 

Let the intensities of the normal com- 
ponents along X y z he an bn Cn respect- 
ively, and the intensities of the pairs of 
tangential components which lie in the 
planes which intersect in x y z and are 



IJ^ GRAPHICAL STATICS. 



121 



perpendicular to those axes be at ht ct re- 
spectively, e.g.^ at is the intensity of the 
tangential component on xoy along y, or 
its equal on xoz along z. 



In Fig. 14 let a plane parallel to the 
given plane cut the axes at x^y^z^\ then 

the total forces on the area x^y^z^ along 
xyz are respectively: 











y 














y 














/ 




























?i 












.._..^ 




1 ' ^ 






*2 /^ 




dx- 








- - y 


^ 






(/V " 










^'Pi^ 
























ya ;\ > 


• — 




\ 
















. 




X 








"^ 










/ 1 






\ 


y^ 




ji^ •• 


\ 




















\ 


y^ 




— ^ 


\ 




\ 






Fig. 14 : 


\ 


/ 1 






/-< 


y^ 


/ 




.^ \ 


1 


^ 




Tx 




c\ 


_y__L__ 




e^/^ 


-^ 


\ 




•A ' ^ 


"a 




""^ 






&i / i 










p..,^" 


_..,... — _ .... ^ 


. \ '■• 


L _ _^ 


' y 


/ ^--1^ ... f. _ 


* 


~ V 




"-"^l^^^^ 


3 


^ 












/ \^^^ ..,!:=' 


yy'- 




=^ 





3?- 

-■V 


-+4 — 


'-^ 










1/ 


iC^^^ \ 


/ ^<#^0l«^*°**°°^ 












1 ' 




' 










1 ^ — --!^ 


i^ 
















1 




z 






^ 


V^>-~^/' 2 












o,| Oj 




;«2 

1 


:e 








V^^ 




-fOt^ 


^-""'^''i^ 


» 1 /'''^\ 








^^~v^ 










] 
































""^ 


\\ 


/^ \ 




/ \ 


\ 






^**'*».. 


.^^^ 






' 




\ 


\ 


I y^^- 


^^.^ 


/ \ 


\ 












^ 








N, y 


>^ " ' " 




-1 \ 
















^y 






V^- 


\ 






\ , 




















< 


s 


z 


a- 


«x 




\ 


\ 


\ 


% 









•^TyA-^i=yi<^2i • ^« + x^oy^ . at + z^ox^.hn 

in which ajb^c^ are the intensities of the 
components of the stress on the plane 
xj/^z^ along xyz respectively. Now 



y^oz^—x^y^z^^ 0,0^ xoi 
z^ox ^—x^y^z ^=^co% yn 
x^oy^ -f- x^y^z^ = cos zn. 

.'. a^=an cos X7i + ht . cos zn + ct cos yn 
h^-=.ct cos jvn + at • cos 2;?i + h^ cos yn 
tf J = hi cos .^n + Cn . cos 2?^ + at cos y?i 

and r''=a3' + 6^' + c/, therefore the result- 
ant stress T is the diagonal of the right 
parallelopiped whose edges are afi^c^. 
In order to construct aj)^c^ it is only 
necessary to lay off an hn Cn, at bt Ct along 
the normal, and take the sums of such 
projections along xyz as are indicated in 
the above values of af>^c^. 

Thus, in Fig. 14, let ^^y^^^ be the 
traces of a plane, and it is required to 
construct the stress upon a plane parallel 
to it through o. 



The ground line between the planes of 
xoy and xoz is ox. The planes xoz and 
yoz on being revolved about ox and oy 
respectively, as in ordinary descriptive 
geometry, leave oz in two revolved posi- 
tions at right angles to each other. 

The three projections of the normal 
at to the given plane are, as is well 
known, perpendicular to the traces of the 
given plane, and they are so represented. 
Let oay be the projection of the normal 
on xoy^ and oay that on xoz. To find 
the true length of the normal, revolve it 
about one projection, say about oa^, and 
if «2 an = <^2 cty then is oan the revolved 
position of the normal. 

Upon the normal let oan = ^n, ohn = 
hn^ ocn = Cn, the given normal compo- 
nents of the stresses upon the rectangu- 
lar planes, and also let oat=at, obt — bt, 
oct = Ct^ the given tangential compo- 
nents upon the same planes. 

Let a.^b^c,^, ^i'^/^-2 ^® ^^^ respective 
projections of the points an bn Cn, dt bt Ct 
of the normal upon the plane xoy by 
lines parallel to 02, similarly ay^ etc., are 
projections by parallels to oy, and ax\ 
etc., by parallels to ox. 

We have taken the stresses Cn and Ct of 



122 



INTERNAL STRESS IN GRAPHICAL STATICS. 



different sign from the others, and so 
have called them negative and the others 
positive. 

It is readily seen that the first of the 
above*- equations is constructed as fol- 
lows: 

a^ = oa^=oa^-\-ht hz' — CzC^' 

Similarly, the other two equations be- 
come: 

h^ = oh^= — oc^ -\'at a^'-\-ob,_ 
c^ = oc^ = ab„' — CzCt + oa^' 

^ We have thus found the coordinates 
01 the extremity r of the stress or upon 
the given plane; hence its projections 
upon the planes of refererence are re- 
spectively OTx, OVy^ Ol'z. 

Problem 28. — In any state of stress 
defined by its three principal stresses, 
to find the stress on any given plane. 

This problem is the special case of 
Problem 27, in which the tangential com- 
ponents are each zero. Taking the nor- 
mal components given in Fig. ,14 as 
principal stresses we find oa^^a^cos xn^ 
oh^^hnGO^ yn^ oc^^= Cn cos zn, as the co- 
ordinates which determine the stress o?*' 
upon the given plane, and the projections 
of or' are ovx, ovy, orz\ respectively. 

From these results it is easy to show 
that the sum of the normal components 
of the stresses on any three planes is 
constant and equal to the sum of the 
principal stresses. This is a general 
property of solid stress in addition to 
those previously stated. 

Problem 29.— Any state of stress be- 
ing defined by given simple stresses, to 
find the stresses on three planes at right 
angles to each other. 

In Fig. 14 let a simple stress act along 
the normal to the plane x^y^z^, and cause 



a stress on that plane whose intensity is 
an = oan, then is a^cos xn=oa^ the in- 
tensity of the stress in the same direction 
acting on the plane yoz. The normal 
component of this latter intensity is 



a. 



cos x7i=:oa^. cos xn=oa^, 



and it is obtained by making oa/-'.oa^y 
a^'ttz" II .Tj^/j, and Uz'Ui^ \\ oy. The tan- 
gential component on yoz is od' in mag- 
nitude and direction, and it is obtained 
thus: make az"d^:^az"ci^\ then in the 
right angled triangle da^a" ^ da^ is the 
magnitude of the tangential component; 
now make od':=da^. This tangential 
component can be resolved along the 
axes of y and z. The stress on the 
planes zox and xoy can be found in simi- 
lar manner, since the tangential compon- 
ents w^hich act on two planes at right 
angles to each other and in a direction 
perpendicular to their intersection are, 
as has been shown, equal; the complete 
construction will itself afford a test of its 
accuracy. 

Other simple stresses may be treated in 
the same manner, and the resultant stress 
on either of the three planes, due to these 
simple stresses, is found by combining 
together the components which act on 
that plane due to each of the simple 
stresses. 

It is useless to make the complete 
combination. It is sufficient to take the 
algebraic sum of the normal components 
acting on the plane, and then the alge- 
braic sum of the tangential components 
along two directions in the plane which 
are at right angles, as along y and z in 
yoz. 

The treatment of conjugate stresses in 
general appears to be too complicated to 
be practically useful, and we shall not 
at present construct the problems arising 
in its treatment. 



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Stoney on Strains. 

New and Revised Edition, with numerous illustrations. Royal 8vo, 664 pp. 

Cloth. $12.50. 

The Theory of Strains in Girders — and Similar Structures, with 
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B. A. 

'" Henrici's Skeleton Structures. 

8vo. Ootli. $1.60. 

Skeleton Structures, especially in their Application to the building 
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Burgli's Modern Marine Engineering. 

One thick 4to voL aoth. $25.00. Half morocco. $30.00. 
Modern Marine Engineering, applied to Paddle and Screw Propul- 
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& Sons; Messrs. Maudslay, Sons & Field ; Messrs. James Watt & Co. ; 
Messrs. J. & G. Rennie ; Messrs. R. Napier & Sons ; Messrs. J. & W. 
Dudgeon ; Messrs. Ravenhill & Hodgson ; Messrs. Ilumplireys & Ten- 
ant ; Mr. J. T. Spencer, and Messrs. Forrester & Co. By N. P. BuRGn> 
Engineer. 

King's Notes on Steam. 

Nineteenth Edition. 8vo. $2.0a 
Lessons and Practical Kotes on Steam, — ^the Steam Engine, Propel- 
lers, &c., &c., for Young Engineers. By the late W. R. Kino, U. S. K. 
Revised by Chief-Engineer J. W. King, U. S. Navy. 



Link and Valve Motions, by W. S. * 
Ancliincloss. 

Sixth Edition. Syo. Cloth. $3.00. 
Application of the Slide Valve and Link Motion to Stationary, 
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tions of the link determined without the assistance of a model. With 
37 wood-cuts and 21 lithogi'aphic plates, with copperplate engraving of 
the Travel Scale. 



Bacon's Steam-Engine Indicator. 

12mo. aoth. $1.00 Mor. $1.60. 

A Treatise on the Richards Steam-Engine IndXcator, — with 
directions for its use. By Charles T. Porter. Revised, with notes 
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Bacon, 1^1. E., Illustrated. Second Edition. 



Islier^^ood's Engineering Precedents. 

Two Vols, in One. 8vo. Cloth. $2.50. 

Engineering Precedents tor Steam Machinery. — By B. F. Isher- 
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Slide Valve by Eccentrics, by Prof. O. W. Mac- 

Gord.. 

4to. Illustrated. Cloth, $3.00 
A Practical Treatise on the Slide Valve by Eccentrics, — 

examining by methods the action of the Eccentric upon the Slide 
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adapting the valve for its various duties in the steam-engine. For the • 
use of Engineers, Draughtsmen, Machinists, and Students of valve 
motions ia general. By C. "VV. MacCord, A. M., Professor of 
Mechanical Drawing, Stevens' Institute of Technology, Hoboken, N. J. 



Stillman's Steam- Engine Indicator. 

li2mo. Cloth. $1,00 
The Steam-Engixe Indicator, — and the Improved Manometer Steam 
and Vacuum Gauges ; their utility and application. By Paul Still- 
man. New edition. 



Porter's Steam-Engine Indicator. 

Tliird Edition. Revised and Enlarged. 8vo. Illustrated. Cloth. $3.50. 
A Treatise on the Kichards Steam-Engine Indicator, — and the 
Development and Application of Force in the Steam-Engine. By 
Charles T. Porter. 



- McCnllocli's Theory of Heat. 

8vo. aoth. 3.50. 
A Treatise on the Mechanical Theory of Heat, and its 
Applications to the Steam-Engine. By Prof. R. S. McCulloch, 
of the Washington and Lee University, Lexington, Va. 



Van Bnren's Formulas. 

8vo. aoth. $2.00. 

Investigations of Formulas, — for the Strength of the Iron parts of 
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Stnart's Snccessfnl Engineer. 

ISmo. Boards. 50 cents. 

IIow to Become a Successful Engineer. Being Hints to Youths 
intending to adopt the Profession. By Bernard Stuart, Engineer. 
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Stuart's Naval Dry Docks. 

Twenty-four engravings on steeL Fourth edition. 4to. Cloth. $6.00. 

TiiK Naval Dry Docks of the United States, By Charles B. 

Stuart, Engineer in Chief U. S. Navy. 



^Vard's Steam for tlie Million. 

8vo. Cloth. $1.0a 

S'IEAm for the Million. A Popular Treatise on Steam and its 
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Ward, Commander U. S. Navy. 



Tnnner on Roll-Tu.rning. 

1 vol. 8vo. and 1 vol. folio plates. $10.00. 
A Treatise on Roll-Turning for the Manufacture of Iron, 
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Grrnner on Steel. 

8vo. Cloth. $3.50. 

The Manufacture of Steel. By M. L. Gruner ; translated from 
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Barba on the Use of Steel. 

12ma Illustrated. Cloth. $1.50. 
The Use of Steel in Construction. Methods of Working, Apply 
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IIOLLEY, P.B. 

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D. VAN NOSTRAND. 



The Useful Metals and tlieir Alloys ; ScofFren, 
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Fifth Edition. 8vo. Half calf. $3.75 
Toe Useful Metals and their Alloys, employed in the conver- 
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William Fairbairn, W. C. Aitkin, and William Vose Pickett. 



Collins' Useful Alloys. 

ISmo. Flexible. 50 cents. 
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Joynson's Metal Used in Construction. 

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Treatise on Ore Deposits. By Bernhard Von Cotta, Professor 
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Plattner's Blow-Pipe Analysis. 

Third Edition. Revised. 568 pages. 8vo. Cloth. $5.00. 

Plattner's Manual of Qualitative and Quantitative Analy- 
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Plymptoix's ]Blo^\^-Pipe Analysis. 

12mo. aoth. $1.50. 
The Blow-Pipe: A Guide to its Use in the Determination of Salts 
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Pynclioii's Clieinical -Physics. 

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Introduction to Chemical Physics ; Designed for the Use of 
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Eliot and Storer's Qnalitative Clieraical 

Analysis. 

New Edition. Revised. 12mo. Illustrated. CJloth. $1.50. 

A Compendious Manual of Qualitative Chemical Analysis. 

By Charles W. Eliot and Frank H. Storer. Revised, with 

the cooperation of the Authors, by "William IIipley Nichols, 

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Rammelsberg's Cliemical Analysis. 

8vo. Cloth. $2.25. 
Guide to a Course op Quantitative Chemical Analysis, 
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by Examples. By C. F. Rammelsberg. Translated by J. Towler, 
M.D, 

Naqnet's Legal Clieinistrj^. 

Illustrated. 12mo. Cloth. $2.00. 
Legal Chemistry. A Guide to the Detection of Poisons, Falsifica- 
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the Use of Chemists, Physicians, Lawyers , Pharmacists, and Experts. 
Translated, with additions, including a List of Books and Memoirs 
on Toxicology, etc., from the French of A. Naquet. By J. P. 
Battershall, Ph. D., with a Preface by C F. Chandler, Ph. D., 
M.D.,LL.D. 



2>. VAN NOSTBAND, 



Prescott's Proximate Organic Analysis. 

12rao. Cloth. $1.75. 

Outlines of Proximate Organic Analysis, for the Identification, 
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Prescott's Alcoliolic Liqnors. 

12ino. Cloth. Si. 50. 
Chemical Examination of Alcoholic Liquors. — A Manual of the 
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Albert B. Prescott, Professor of Organic and Applied Chemistry 
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Prescott and Douglas's Ctnalitative Cliemi- 

cal Analysis. 

Second Edition. Revised. 8vo. Cloth. $3.50. 
A Guide in the Practical Study of Chemistry and in the Work of Analysis. 



Pope's Modern Practice of tlie Electric 

Telegraph.. 

Ninth Edition. 8vo. Cloth. $2.00. 
A Iland-book for Electricians and Operators. By Frank L. Pope. 
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Sabine's History of tlie Telegrapli. 

Second Edition. 12ino. Cloth. $1.25. 
History and Progress of the Electric Telegraph, with De- 
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Haskins' Gralvanometer. 

Pocket form. Illustrated. Morocco tucks. $2.00. 
The Galvanometer, and its Uses ; — A Manual for Electricianf 
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Larrabee's Secret Letter and Telegraph. 

18mo. Cloth. Sl.OO. 
Cipher and Secret Letter and Telegraphic Code, with Hogg s 
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10 SCIENTIFIC BOOKS PUBLISHED BY 

Gillmore's Limes and Cements. 

Fifth Edition. Revised and Enlarged. 8vo. Cloth. $4.00. 

Practical Treatise ox Limes, Hydraulic Cements, axd Mor- 
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Brevet Major-General U. S. Army. 



Gillmore's Coignet Beton. 

Nine Plates, Views, etc. 8vo. Cloth. $2.50. 

Coignet Beton and Other Artificial Stone. — By Q. A. Gill- 
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Grillmore on Roads. 

Seventy Illustrations. 12mo. Cloth. $2.00. 

A Practical Treatise on the Construction op Roads, Streets, 
AND Pavements. By Q. A. Gillmore, Lt.-Col. U. S. Corps of 
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Grillmore's Bnilding Stones. 

8vo. Cloth. $1.00. 

Report on Strength of the Building Stones in the United 
States, etc. 

Holley's Rail^vay Practice. 

1 vol. folio. Cloth. $12.00. 

American and European Railway Practice, in the Economical 
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burning ; and in Permanent Way, including Road-bed, Sleepers, 
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L. IIoLLEY, B.P. AVith 77 lithographed plates. 



Useful Information for Railway Men. 

Pocket form. Morocco, gilt. $2.00. 

Compiled by W. G. Hamilton, Engineer. Kew Edition, Revised 
and Enlarged. 577 pages. 



D, VAN NOSTRAND. 11 



Stuart's Civil and Military Engineers of 

America. 

8vo. Illustrated, Cloth. $5.00. 
The Civil and Military Engineers of America. By General 
Charles B. Stuart, Author of " Naval Dry Docks of the United 
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of the most important and original works constructed in America. 



Ernst's Manual of Military Engineering. 

193 Wood-cuts and 3 Lithographed Plates. 12nio. Cloth. $5.00.; 
A Manual of Practical Military Engineering. Prepared for 
the use of the Cadets of the U. S. Military Academy, and for Engineer 
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Simms' Levelling. 

12EQO. Cloth. $2.50. 
A Treatise on the Principles and Practice of Levelling, 
showing its application to purposes of Railway Engineering and the 
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Mr. Law's Practical Examples for Setting-out Railway Curves. 
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JelFers' Nantical Surveying. 

Illustrated with 9 Copperplates and 31 Wood-cut Illustrations. 8vo. Cloth. $5.00. 
Nautical Surveying. By William N. Jeffers, Captain U. S. 
Navy. . 

Text-book of Snrveying. 

8vo. 9 Lithograph Plates and several Wood-cuts. Cloth. $3.00. 
A Text-book on Surveying, Projections, and Portable Instruments, 
for the use of the Cadet Midshipmen, at the U. S. Naval Academy. 



The Plane Table. 

8vo. Cloth. $2.00. 

Its Uses in Topographical Surveying. From the papers of the 
U. S. Coast Survey. 



Chaiivenet's Lunar Distances. 

8vo. Cloth. $2.00. 
New Method of Correcting Lunar Distances, and Improved 
Method of Findin-g the Error and Rate of a Chronometer, by equal 
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Bnrt's Key to Solar Compass. 

Second Edition. Pocket-book form. Tuck. $2.60. 
Key to the Solar Compass, and Surveyor's Companion ; comprising 
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Linear Surveys and Public Land System of the United States, Notes 
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Ho^vard's Earth.\^rork: Mensuration. 

8vo. niustrated. Clotli. $1.50. 
Earthwork Mensuration on the Basis op the Prismoidal 
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Prismoidal Contents, directly from End Areas. Illustrated by 
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Conway R. Howard, Civil Engineer, Richmond, Va. 



[Morris' Easy Rules. 

78 Illustrations. 8vo. Cloth. $1.50. 
Easy Rules for the Measurement of Earthworks, by means of 
the Prismoidal Formula. By Elwood Morris, Civil Engineer. 



Clevenger's Surveying. 

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A Treatise on the Method of Government Surveying, as 
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Students who contemplate engaging in the business of Public Land 
Surveying. By S. V. Clevenger, U. S. Deputy Surveyor. 



He'wson on Embankments* 

8vo. aoth. $2.00. 
Principles and Practice of Embanking Lands from River 
Floods, as applied to the Levees of the Mississippi. By William 
Hewson, Civil Engineer. 



Minifie's Mecliaiiical Drawing. 

Ninth Edition. Royal 8vo. Cloth. $4.00. 

A Text-Book of Geometrical Drawing, for the use of Mechanics 
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With over 200 diagrams on steel. By William Minifie, Architect. 
With an Appendix on the Theory and Application of Colors. 



Minifie's Greometrical Or a wine:. 



C3' 



New Edition. Enlarged. 12mo. Cloth. $2.00. 



Geometrical Drawing. Abridged from the octavo edition, for the 
use of Schools. Illustrated with 48 steel plates. 



Free Hand. Drawing. 

Profusely Illustrated. 18mo. Boards. 60 cents. 

A Guide to Ornamental, Figure, and Landscape Drawing. By an 
Art Student. 



Tlie Meclianic's Friend. 

12mo. Cloth. 300 Illustrations. Si. 50. 
The Mechanic's Friend. A Collection of Receipts and Practical 
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graphy — Taxidermy — ^Varnishes — Waterproofing — and Miscellaneous 
Tools, Instruments, Machines, and Processes connected with the 
Chemical and Mechanical Arts. By William E. Axon, M.R.S.L. 



Harrison's Meclianic's Tool-Book:. 

M Illustrations. 12mo. Cloth. $1.50. 
Mechanics' Tool Book, with Practical Rules and Suggestions, for the 
use of Machinists, Ii'on Workers, and others. By W. B. Harrison. 



Randall's Qnartz Operator's Hand-Book. 

12mo. Cloth. $2 00. 

Quartz Operator's Hand-Book. By P. M. Randall. New 

edition. Revised and Enlarged. Fully illustrated. 



14 SCIENTIFIC BOOKS PUBLISHED BY 

Joynson on Machine Q-earing. 

8vo. Cloth. $2.00. 
The Mechanic's and Student's Guide in the designing and Con* 
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Wheels, etc., and the Drawing of Rectilineal and Curved Surfaces. 
Edited by Francis H. Joynson. With 18 folded plates. 



Silversmitli's Hand-Book. 

Fourth Edition. Illustrated. 12nio. Cloth. $3.00. 

A Practical Hand-Book for Miners, Metallurgists, and Assayers. 
By Julius Silversmith. Illustrated. 



Barnes' Submarine Warfare. 

8vo. Cloth. $5.00. 
Submarine Warfare, Defensive and Offensive. Descriptions 
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made tD deterinine the Explosive Force of Gunpowder under Water. 
Also a discussion of the Offensive Torpedo system, its effect upon 
Iron-clad Ship systems, and influence upon future Naval Wars. By 
Lieut.-Com. John S. Barnes, U.S.N. With twenty lithographic 
plates and many wood-cuts. 

Foster's Submarine Blasting. 

4to. Cloth. $3.50. 
Submarine Blasting, in Boston Harbor, Massachusetts — Removal of 
Tower and Corwin Rocks. By John G. Foster, U. S. Eng. and 
Bvt. Major-General U. S . Army. With seven plates. 



Mowbray's Tri-Nitro-Grtycerine. 

8vo. Cloth, niustrated. $3.00. 
Tri-Kitro-Glycerine, as applied in the Hoosac Tunnel, and to Sub- 
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\Villiamson on the Barometer. 

4to. Cloth. $15.00. 
On the Use of the Barometer on Surveys and Reconnais- 
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Part n. — Barometric Hypsometry. By R. S. Williamson, Bvt. 
Lt.-Col. U. S. A., Major Corps of Engineers. With illustrative tables 
and engravings. 



Williamson's Meteorological Tables. 

4to. Flexible Cloth. $2.50. 

Practical Tables in Meteorology and Hypsometry, in connection 
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Butler's Projectiles and Rifled Cannon. 

4to. 36 Plates. Cloth. $7.50. 

Projectiles and Rifled Cannon. A Critical Discussion of the 
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Benet's Clironoscope. 

Second Edition. Illustrated. 4to. Cloth. $3.00. 

Electro-Ballistic Machines, and the Schultz Chronoscope. By 
Lt.-Col. S. V. Benet, Chief of Ordnance U. S. A. 



Micliaelis' Chronograph. 

4to. Illustrated. Cloth. $3.00. 

The Le Boulenge Chronograph. With three lithographed folding 
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Nugent on Optics. 

12mo. Cloth. $1.50. 

Treatise on Optics ; or. Light and Sight, theoretically and practically 
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Peirce's Analytic Mechanics. 

4to. Cloth. $10.00. 
System of Analytic Mechanics. By Benjamin Peirce, Pro- 
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Craig's Decimal System. 

Square 32mo. Limp. 60a 

Weights and Measures. An Account of the Decimal System, with 
Tables of Conversion for Commercial and Scientific Uses. By B. P. 
Craig, M.D. 



16 SCIENTIFIC BOOKS PUBLISHED BY 



Alexander's Dictionary of Weights and. 

Measures. 

New Edition. 8vo. Cloth. $3.50. 
Universal Dictioxary of Weights and Measures, Ancient and 
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By J. H. Alexander. 

Elliot's European Liglit-Honses. 

51 Engravings and 21 Wood-cuts. 8vo. Cloth. $5.00. 
European Light-House Systems. Being a Report of a Tour of 
Inspection made iu 1873. By Major GitORGE H. Elliot, U. S. 
Engineers. 

Sweet's Keport on Coal. 

With Maps. 8vo. Cloth. $3.00. 
Special Report on Coal. By S. H. Sweet. 



Colbnrn's Gras Works of London. 

12mo. Boards. 60 cents. 
Gas Works of London. By 2Jerah Colburn. 

Walker's Scre^v Propulsion. 

Svo. Cloth. 75 cents. 

Notes on Screw Propulsion, its Rise and History. By Capb. W. H. 
Walker, U. S. Navy. 

Pook on Sliipbnilding. 

Svo. Cloth. Illustrated. $5.00. 
Method of Preparing the Lines and Draughting Vessels 
Propelled by Sail or Steam, including a Chapter on Laying-off 
on the Mould-loft Floor. By Samuel M. Pook, Naval Constructor. 

Saeltzer's Acoustics. 

12mo. Cloth. $2.00. 
Treatise on Acoustics in connection with Ventilation. By Alex- 
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Eassie on Wood and its XTses. 

250 Illustrations. Svo. Cloth. $1.50. 

A Hand-book for the Use op Contractors, Builders, Architects, 
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Wanklyn's Milk Analysis. 

12mo. Cloth. $1.00. 

Milk Analysis. A Practical Treatise on the Examination of Milk, 
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AVanklyn,M.R.C.S. 



Rice & Jolmsoii's OifFerential Fanctions. 

Paper, I'imo. 50 cents. 

Ox A New Method of Obtaining the Differentials of Func- 
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and W. WooLSEY Johnson, Prof, of Mathematics, St. John's 
College, Annapolis. 



Coffin's Navigation. 

Fifth Edition. 12mo. Qoth. $3.50. 



Navigation and Nautical Astronomy. Prepared for the use of 
the U. S. Naval Academy. By J. H. C. Coffin, Professor of 
Astronomy, Navigation and Surveying ; with 52 wood-cut illustra- 
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Clark's Theoretical Navigation, 

8vo. Qoth. $3.00. 

Theoretical Navigation and Nautical Astronomy. By Lewis 
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Toner's Dictionary of Elevations. 

8vo. Paper, S3.00 Cloth, S3, 75. 

Dictionary of Elevations and Climatic Register of the 
United States. Containing, in addition to Elevations, the Latitude, 
Mean Annual Temperature, and the total Annual Kain Fall of many 
Localities ; with a brief introduction on the Orographic and Physical 
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18 



SCIENTIFIC BOOKS PUBLISHED BY 



VAN NOSTKAND'S SCIENCE SEMES. 



It is the intention of the Publisher of this Series to issue them at 
intervals of about a month. They will be put up in a uniform, neat, 
and attractive form, 18mo, fancy boards. The subjects will be of an 
eminently scientific character, and embrace as wide a range of topics as 
possible, all of the highest character. 

Price, 50 Ceuts Eacli. 

I. Chimneys for Furxaces, Fire-places, and Steam Boilers. By 
R. Armstrong, C.E. 

ir. Steam Boiler Explosions. By Zerah Colburn. 

III. Practical Designing of Retaining Walls. By Arthur Jacob, 
A. B . With Illustrations. 

lY. Proportions of Pins Used in Bridges. By Charles E. 
Bender, C.E. With Illustrations. 

V. Ventilation OF Buildings. By W. F. Butler. With Illustrations. 

YI. On the Designing and Construction of Storage Reservoirs. 
By Arthur Jacob. AVith Illustrations. 

YII. Surcharged and Different Forms of Retaining Walls. 
By James S. Tate, C.E. 

YIII. A Treatise on the Compound Engine. By John Turnbull. 

With Illustrations. 

IX. Fuel. By C.William Siemens, to which is appended the value of 
Artificial Fuels as Compared with Coal. By John Worm" 
ALD, C.E. 

X. Compound Engines. Translated from the French of A. Mallet. 

Illustrated. 



XI. Theory of Arches. By Prof. W. Allan, of the Washington and 

Lee College. Illustrated. 

XII A Practical Theory of Youssoir Arches. By William Cain, 
C.E. Illustrated. 



D. VAN NOSTBAND. 19 



XIII. A Practical Treatise ox the Gases Met With in Coal 
Mines. By the late J. J. Atkinson, Government Inspector of 

Mines for the County of Durham, England. 

XIV. Friction of Air in Mines. By J. J. Atkinson, author of " A 
Practical Treatise on the Gases met with in Coal Mines." 

XV. Skew Arches. By Prof. E. W. Hyde, C.E. Illustrated with 
numerous engravings and three folded plates. 

XVI. A Graphic Method for Solving Certain Algebraic Equa- 
tions. By Prof. George L. Vose. With Illustrations. 

XVII. Water and Water Supply. By Prof. W. II. Corfield, 
M.A., of the University College, London. 

XVIII. Sewerage and Sewage Utilization. By Prof. W. H. 
Corfield, M.A., of the University College, London. 

XIX. Strength of Beams Under Transverse Loads. By Prof. 
W. Allan, author of "Theory of Arches." With Illustrations 

XX. Bridge and Tunnel Centres. By John B. McMasters, 
C.E. With Illustrations. 

XXI. Safety Valves. By Richard II. Buel, C.E. With Illustra- 
tions. 

XXn. High Masonry Dams. By John B. McMasters, C.E. 

With Illustrations. 

XXm. The Fatigue of Metals under Repeated Strains, with 
various Tables of Results of Experiments. From the German of 
Prof. Ludwig Spangenberg. With a Preface by S. H. Shreve, 
A.M. With Illustrations. 

XXIV. A Practical Treatise on the Teeth of Wheels, with 
the theory of the use of Robinson's Odontograph. By S; W. Robin- 
son, Prof, of Mechanical Engineering, Illinois Industrial University. 

XXV. Theory and Calculations of Continuous Bridges. By 
Mansfield Merriman, C.E. With Illustrations. 

XXVI. Practical Treatise on the Properties of Continuous 
Bridges. By Charles Bender, C.E. 



20 SCIENTIFIC BOOKS BUBLISHED BY 



XXVII. On Boidbr Inckustation and Corrosion, By J, F. Rowan. 

XXVIII. On Transmission op Power by Wire Rope. By Albert "W. 
Stabl. 

XXIX. Injectors : Their Theory and Use. Translated from the 
French of M. Leon Pouchet. 

XXX. Terrestrial Magnetism and the Magnetism op Iron Ships. 
By Professor Fairman Rogers. 

XXXI. The Sanitary Condition of Dwelling Houses in Town and 
Country. By George E. "Waring, Jr. 



IN PRESS, 



Heating and Ventilation in its Practical Ap- 
plication for tlie Use of Engineers and 
Architects. 

Embracing a Series of Tables and Formulae for dimensions for Heating 
Flow and Return Pipes, for Steam and Hot Water Boilers, Flues, etc., 
etc. By F. Schumann, C. E. 1 vol. 13mo. Illustrated. 



A Griiide to th.e Determination of Rocks. 

Being an Introduction to Litholagy. By Edward Jannettaz, Doctuer des 
Sciences. Translated from the French by Geo. W. Plympton, Profes- 
sor of Physical Science, Brooklyn Polytechnic Institute. 12mo. 



Shield's Treatise on Enigineering 
Constrnction, 

12mo. Cloth. 

Embracing Discussions of the Principles involved and Descriptions of the 
Material employed in Tunnelling, Bridging, Canal and Road Build- 
ing, etc., etc. 



r' 



MILlTAllY BOOKS 

PUBLISHED EY 

D. VAN NOSTRAND, 

23 Murray Street and 27 Warren Street, 
NEWYOBK. 



Any Book in this Catalogue sent free by mail on receipt of price. 



Benton's Ordnance and G-nnnery. 

Fourth Edition, Revised and Enlarged. 8vo. Clotli. .S5.00. 
Ordnance and Gunnery. A Courso of Instruction iu Ordnanco 
and G unnery. Compiled fortlio iiS3 of tho Cadots of tlio U. S. Military 
Academy, by Col. J. G. Benton, Major Ordnanco Dcp., late Instructor 
of Ordnance aud Gunnery, Military Academy, West Point. Illus- 
trated. 

Holley's Ordnance and Armor. 

Svo. Half Roan, $10.00. Half Russia, $12.00. 
A Treatise on Ordnance and Armor. With an Appendix, refer- 
ring to Oun-Cotton, Hooped Guns, etc. , etc. By Alexander L. lloUey, 
B. P. With 493 illustrations. 948 pages. 

Scott's Military Dictionary. 

8x0. Half Roan, SOvOO. Half Russia, ^8.00. Full Morocco, $10.00. 
Military , Dictionary. Comprising Technical Definitions ; Informa- 
tion on Raising and Keeping Troops ; Law, Government, Kegu- 
lation, and Administration relating to Land Forces. By Col. II. L. 
Scott, U.S.A. 1vol. Fully illustrated. 



Koemer's Cavalry. 

Svo. Cloth, $0.00. Half Calf, $7.50. 
Cavalry : Its History, Management, and Uses in War. By J. 
Koemer, LL.D., late an officer of Cavalry in the Service of the Nether- 
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wood engravings. Beautifully printed on tinted paper. 

25 



2G MILITAIiY BOOKS PUBLISHED BY 



Micliaelis' Clironograpli. 

4to. Illustrated. Cloth. $3.00, 
The Le Boulexge Ciiroxograph. With three lithographed folding 
plates of illustrations. By Brevet Capt. O. E. Michaelis, First Lieu- 
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IBenet's Cliroiioscope. 

Second Edition. Illustrated. 4to. Cloth. $3.00. 
Electro-Ballistic Machines., and the Schultz Chronoscope. By 
GenL S. Y. Benct, Chief of Ordnance, U. S. Army. 

DnfoTir^s Principles of Strategy and. GrrancI 

Tactics. ^ — 

12mo. Cloth. S3.00. 
The Prixciples of Strategy ani> Grand Tactics. Translated 
from the French of General G. II. Dufour. By William P. Craighill, 
U. S. Engr. , and late Assistant Professor of Engineering, Military 
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Jomini's Life of tlao Emperor jSTapoleon. 

4 vols. 8vo., and Atlas. Cloth. Half Calf, 

Military and Political Life of the Emperor Kapoleon. By 
Baron Jomini, General-in-Chief and Aid-de-Camp to tlie Emperor cf 
Hussia. Translated from the French, with IsTotes, by II.. Yf. Halleekj 
LL.D., Major-Generai U. S. Army. With GO l^Iaps and Plans. 



Jominfs Campaign of Waterloo. 

Tliird Edition. 12hio. Cloth, s^l.25. 
The Political and Military History of Tiiii: Campaign of Y^a- 
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JoiTiini's Grranci Military Operations. 

2vol.s.8vo,, r.ud Atlas. Qoth, $15.00. Half Calf or Morocco, S21. Half Russia, 

Treatise on Grand Military Operations. Illustrated by a Critical 
and Military History of the "\7ars of Frederick the Great. Y/ith a 
Summary of the ^lost Important Principles of the Art of War. By 
Baron de Jomini. Illustrated by Maps and Plans. Translated froiri 
the French by Col. S. B. Ilolabird, A. I). C, U. S. Army. 



J). VAN NOSTRAND. 



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Royal 8vo. Illustrated with Cliromo-Lithographs. Extra Cloth. $7,50. 
Everglade to Canon, with the Second Dragoons (Second U. S. Car- 
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the Indian Country, including Personal Hecollections of Distinguished 
Officers. By Theo. F. Eodenbough, Colonel and Brevet Brigadier- 
General, U. S. Army. 

History of Brevets. 

Crown 8vo. Extra Cloth, $3.50. 
The History and Legal, Effects of Bkevets in Iho Armies of 
Great Britain and the United States, from the origin in 1C02 until the 
present time. By Gen. James B. Fry, U. S. Army. 

Barre Dnparcq's Military Art and llistory. 

8vo. Cloth. $5.00. 
Elements of Military Art and History. By Edward de la Barre 
Duparcq, Chef de Bataillon of Engineers in the Army of France, and 
Professor of the Military Art in the Imperial School of St. Cyr. 
Translated bv Colonel Geo. AV. CuUum, U. S. E. 



Discipline and Drill of tbe Militia. 

Crown 8vo. Flexible cloth. $3.00. 

The Discipline and Drill of the Militia. By Major Frank S. 
Arnold, Assistant Quartermaster-General, Rhode Island. 



^Vallen's Service Man^ial. 

12mo. Cloth. $1.50. 
Service Manual for the Instruction of newly appointed Commissioned 
Oincers, and the Rank and File of the Army, as compiled from Army 
Regulations, The Articles of War, and the Customs of Service. By 
Henry D. Yf alien, Bvt. Brigadier-General U. S. Army. 



Boynton's I-listory of West Point. 

Second Edition, 8^'o. Fancy Cloth. $3.50. 
History of V/'est Point*, and its Military Importance during the 
American Revolution ; and the Origin and Progress of the United 
States Military Academy. By Bvt. Maj. Edward C. Boynton, A. M., 
Adjutant of the I^Iilitary Academy. With oG Maps and Engraving 



28 MILITABY BOOKS PUBLISHED BY 

\¥ood's West Point Scrap-Book. 

8vo, Extra Cloth. $5,00 
The West Point Scrap-Book. Being a Collection of Legends, Stories, 
Songs, &c. By Lieut. O. E. Wood, U. S. A. With 69 wood-cut 
Illustrations. Beautifully printed on tinted paper. 

West Point Life. 

Oblong 8vo, Cloth, $2.50, 
West Point Life. A Poem read before the Dialectic Society of the 
United States Military Academy. Illustrated with twenty-two full- 
page Pen and Ink Sketches. By A Cadet. To which is added tha 
song, "Benny Havens, Oh!" 



Grillmore's Fort Snmter. 

8vo. Cloth. $10,00. Half Russia, $12.00. 
Gillmore's Fort Sumter. Official Report of Operations against the 
Defences of Charleston Harbor, 1863. Comprising the descent upon 
Morris Island, the Demolition of Fort Sumter, and the siege and 
reduction of Forts Wagner and Gregg. By Ma j. -Gen. Q. A. Gill- 
more, U. S. Engineers. With 76 lithographic plates, views, maps, etc. 



Gillmore's Supplementary Report on Fort 

Snmter. 

8vo. Cloth. $5.00, 
Supplementary Report to the Engineer and Artillery Operations 
against the Defences of Charleston Harbor in 1863. By Maj.-Gen. Q. 
A. Gillmore, U. S, Engineers. With Seven Lithographed Maps and 
Views. 



Grillmore's Fort Pulaski. 

8vo, Cloth, $2.50 

Siege and Reduction of Fort Pulaski, Georgia. By Maj.-Gen. 

Q. A. Gillmore, U. S. Engineers. Illustrated by Maps and Views. 



Barnard and Barry's Report. 

8vo. Cloth, S4.00. 
Report of the Engineer and Artillery Operations of the 
Army of the Potomac, from its Organization to the Close of the 
Peninsular Campaign. By Maj.-Gen. J. G. Barnard, U. S. Engineers, 
and Maj.-Gen. W. F. Barry, Chief of Artillery. Illustrated by 18 
Maps, Plans, &c. 



D, VAN NO STRAND. 



29 



Griiide to ^Vest Point. 

18mo. Flexible Cloth. $1,00, 
Guide to West Poixt axd the IJ. S. Military Academy. With 
Maps and Engravings. 



Barnard's C. S. A., and tlie Battle of Bull 

Ilnn. 

8vo. Cloth. $2.00, 
The "C. S. A.," and the Battle of Bull Run. By Maj.-Gen. J. G. 
Barnard, U. S. Engineers. With five Maps. 



Barnard's Peninsnlar Campaign. 

8vo, Cloth. $1.00. 12mo. Paper 30c. 
The Peninsular Campaign and its Antecedents, as developed by 
the Report of Maj.-Gen. Geo. B. McClellan, and other published 
Documents. By Maj.-Gen. J. G. Barnard, U. S. Engineers. 



Barnard's Notes on Sea-Coast Defence. 

8vo. Cloth. $2.00. 
Notes on Sea-Coast Defence : Consisting of Sea-Coast Fortifica- 
tion ; the Fifteen-Inch Gun ; and Casemate Embrasure. By Major- 
Gen. J. G. Barnard, U. S. Engineers. AVith an engraved Plate of 
the 15-inch Gun. 



Henry's Military Record of Civilian 
Appointments, U. S. A. 

2 Vols. Svo. Cloth. SlO.OO. 

Military Record of Civilian Appointments in the United 
States Army. By Guy V. Henry, Brevet-Colonel U. S. A. 



Harrison's Pickett's Men. 

12mo. Cloth. S2.00. 
Pickett's Men. A Fragment of War History. By Col. Walter Har- 
rison. W^ith portrait of Gen. Pickett. 



Todleben's Defence of Sebastopol. 

12mo. Cloth. $2.00. 
Todleben's (General) History of the Defence of Sebastopol. 
By William Howard Russell, LL.D., of the London Times. 



so 



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Hotclikiss and Allan's Battle of Cliaiieellors- 

ville. 

8vo. Cloth. ^5.00. 
The Battle-fields of Virginia. ChancellorsVille, embracing the 
Operations of the Army of Northern Virginia. From the First Battle 
of Fredericksburg to the Death of Lt.-Gen. T. J. Jackson. By Jed. 
Ilotchkiss and AVilliam Allan. Illustrated with five Maps and Por- 
trait of Stonewall Jackson. 



Andre^vs' Campaign of Mobile. 

8vo. Cloth. $3.50. 
The Campaign of Mobile, including the Co-operation of General 
Wilson's Cavalry in Alabama. By Brevet Maj.-Gen. C. C. Andrews. 
With five Maps and Views. 



Stevens' Three Years in tlie Sixth. Corps. 

N'ew and Revised Editiou. 8vo. Cloth. $3.00 

Three Years in the Sixth Corps. A concise narrative of events in 

the Army of the Potomac from 1861 to the Close of the Bebellion. 

April, 1865. By Geo. T. Stevens, Surgeon of the 77th Regt. New 

York Volunteers. Illustrated with 17 engravings and six steel portraits. 

Lecomte's War in the United States. 

12mo. Cloth. $1.00. 
The AVar in the United States. A Report to the Swiss Military 
Department. By Ferdinand Lecomte, Lieut.-Col. Swiss Confedera- 
tion. Translated from the French by a Staff Officer. 



Koberts' Hand-Book of Artillery. 

IGmo. Morocco Clasp. $2.00. 
IIand-Book of Artillery. For the service of the United States 
Army and Militia. Tenth edition, revised and greatly enlarged. By 
Joseph Roberts, Lt.-Col. 4th Artillery and Brevet. Maj.- General U. S. 
Army. 



Instrnctions for Field Artillery. 

12rao. Cloth. $3.00. 
Instructions for Field Artillery. Prepared by a Board of Artil- 
lery Officers. To which is added the "Evolutions of Batteries," 
translated from the French, by Brig.-Gen. R. Anderson, U. S. A. 122 
plates. 



D. VAN NO STRAND, 



31 



Hea^^y Artillery Tactics. 

12rao. Cloth. S2.50. 
Heavy Artillery Tactics. — 1863. Instructions for Heavy Artillery; 
prepared by a Board of Officers, for the use of the Army of the United 
States. With service of a gun mounted on an iron carriage and 3.9 
plates. 



Andersons' Evolutions of Field Artillery. 

24mo. Cloth. $1.00. 
Evolutions of Field Batteries of Artillery. Translated from 
the French, and arranged for the Army and Militia of the United 
States. By Gen. llobert Anderson, U. S. A. Published by order of 
the War Department. 33 plates. 



Dnane's Manual for Engineering- Troops. 

12nio. Half Morocco. ^2.50. 
Manual for Engineer Troops : Consisting of — Part I."' Ponton Drill ; 
II. Practical Operations of a Siege; III. School of the Sap ; IV. Mili- 
tary Mining ; V. Construction of Batteries. By General J. C. Duane, 
Corj)s of Engineers, U. S. Army. With IG pla,tes and numerous wood- 
cut illustrations. 



CLillnm's Military Bridges. 

8vo. Cloth. $3.50. 
Systems of Military Bridges, in use by tho United States Army; 
those adopted by the Great European Powei s ; and such as are em- 
ployed in British India. With Directions for the Preservation, 
Destruction', and Re-establishment of Bridges. By Col. George Y>^. 
' Cullum, U. S. E. With 7 folding plates. 



Mendell's Militar^^ Surveying. 

12m o. Cloth. S2.00. 
A Treatise on Military Surveying. Theoretical and Practical, 
including a description of Surveying Instruments. By G. II. Mendell, 
Major of Engineers. With 70 wood-cut illustrations. 



Abbot's Siege Artillery Against Kidinaond. 

8vo. Cloth. $3.."0. 
Siege Artillery in the Campaign Against Biciimond. 
L. Abbot, Major of U. S. Engineers. Illustrated. 



By Ilcnry 



32 MILITARY BOOKS PUBLISHED BY 

Haupt's Military Bridges. 

8vo. Cloth. $6.50. 
Military Bridges ; For the Passage of Infantry, Artillery and Bag- 
gage Trains ; with suggestions of many new expedients and construc- 
tions for crossing streams and chasms. Including also designs for 
Trestle and Truss-Bridges for Military Railroads, adapted specially to 
the wants of the Service of the United States. By Herman Haupt, 
Brig.-Gen. U. S. A., author of *' General Theory of Bridge Construc- 
tions," &c. Illustrated by 69 lithographic engravings. 



Lendy's Maxims and Instructions on the 

Art of War. 

18mo. Cloth. 75c. 
Maxims and Instructions on the Art of "War. A Practical 
Military Guide for the use of Soldiers of All Arms and of all Coun- 
tries, Translated from the French by Captain Lendy, Director of the 
Practical Military College, late of the French Staff, etc., etc. 



Benet's Military La^v and Conrts-Martial."' 

Sixth Editiou, Revised and Enlarged. 8vo. Law Sheep. $4.50- v.;^ J 
Benet's Military Law. A Treatise on Military Law and the Prac- 
tice of Courts-Martial. By Gen. S. V. Benet, Chief of Ordnance U. S. A., 
late Assistant Professor of Ethics, Law, &c., Military Academy, West 
Point. 



Lippitt's Special Operations of War, 

Illustrated. 18mo. Cloth. $1.00. 



Lippitt's Field Service in War. 

12mo. Cloth. ■ $1.00. 



Lippitt's Tactical Use of tlie Three Arms. 

12mo. Cloth. $1.00. 



Lippitt on Intrenclinients. 

41 Engravings. 12mo. Cloth. $1.25. 



Kelton's Nevi^ Bayonet Exercise, 

Fifth Edition. Revised. 12mo. Cloth. $2.00. 
New Bayonet Exercise. A New Manual of the Bayonet, for the 
Army and Militia of the United States. By General J. C. Kelton, 
U. S. A. "With 40 beautifully engraved plates. 



Craighiirs Army Officers' Companion. 

18mo. Full Roan. $2.00. 
The Army Officers' Pocket Companion. Principally designed for 
Staff Officers in the Field. Partly translated from the French of 
M. de Rouvre, Lieut.-Col. of the French Staff Corps, with additions 
from Standard American, French, and English authorities. By Wm. 
P. Craighill, Major U. S. Corps of Engineers, late Assistant Professor 
of Engineering at the U. S. Military Academy, West Point. 

Casey's U. S. Infantry Tactics. • 

3 vols. 24mo. Cloth. $2.50. 
U. S. Infantry Tactics. By Brig.-Gen. Silas Casey, U. S. A. 3 vols., 
24mo. Vol. I. — School of the Soldier; School of the Company; In- 
struction for Skirmishers. Vol. II. — School of the Battalion. Vol. 
HI. — Evolutions of a Brigade ; Evolutions of a Corps d'Armee. 
Lithographed plates. 

United States Tactics for Colored Troops. 

24mo. Cloth. $1.50. 
U. S. Tactics for Colored Troops. U. S. Infantry Tactics for the 
use of the Colored Troops of the United States Infantry. Prepared 
under the direction of the War Department. 



Morris' Field Tactics for Infantry. 

Illustrated. 18mo. Cloth. 75c. 
Field Tactics for Infantry. By Brig.-Gen. Wm. II. Morris, U. S. 
Vols., late Second U. S. Infantry. 

Monroe's Liglit Infantry and Company Drill. 

32mo. Cloth. 75c. 
Light Infantry Company and Skirmish Drill. Bayonet Fencing ; 
with a Supplement on the Handling and Service of Light Infantry. 
By J. Monroe, Col. Twenty- Second Regiment, N. G., N. Y. S. M. for- 
merly Captain U. S. Infantry. 

Berriman's Sword Play. 

Fourth Edition. 12mo. Cloth. Si- 00. 
Sword-Play. The Militiaman's Manual and Sword-Play without a 
Master. Bapier and Broad- Sword Exercises, copiously explained and 
illustrated; Small- Arm Light Infantry Drill of the United States 
Army ; Infantry Manual of Percussion Musket ; Company Drill of the 
United States Cavalry. By Major M. W. Berriman. 



Morris' Infantry Tactics. 

2 vols. 24mo. $2.00. 2 vols, in 1. Cloth. $1.50. 
Infantry Tactics. By Brig. -Gen. William 11. Morris, U. S. Vols, 
and late U. S. Second Ii>M«try. 



8 



Le Oal's Scliool of tlie Grnides. 

16mo. Cloth. 60c. 
The School of the Guides. Designed for the use of the Militia of 
the United States. By Col. Eugene Le Gal. 



Dnryea's Standing Orders of tlie Seventli 

Regiment. 

New Edition. 16mo. Cloth. 50c. 
Standing Orders of the Seventh Regiment National Guards. 
By A. Duryea, Colonel. 



Heth-'s System of Target Practice. 

18mo. Cloth. 75c. 
System of Target Practice ; For the use of Troops when armed 
with the Musket, Rifle-Musket, Rifle, or Carbine. Prepared princi- 
pally from the French, by Captain Henry Heth, Tenth Infantry, 
U. S. A. 



Wilcox's Rifles and Rifle Practice. 

New Edition. Illustrated. 8vo. Cloth. $2.00. 
Rifles and Rifle Practice. An Elementary Treatise on the Theory 
of Rifle Firing ; with descriptions of the Infantry Rifles of Europe 
and the United States, their Balls and Cartridges. By Captain C. M. 
AVilcox, U. S. A. 



Viele's Pland-Book for Active Service. 

12rao. Cloth. $1.00. 
IIand-Book for Active Service, containing Practical Instructions in 
Campaign Duties. For the use of Volunteers. By Brig.-Gen. Egbert 
L. Viele, U. S. A. 



Nolan's System for Training Cavalry Horses. 

24 Plates. Cloth. $2.00. 
ISTolan's System for Training Cavalry Horses. By Kenner Gar- 
rard, Bvt. Brig.-Gen. U. S. A. 



D. VAN NO STRAND. 35 



Arnold's Cavalry Service. 

Illustrated ISino. Cloth. 75c. 
Notes ox Horses for Cavalry Service, embodying the Quality, 
Purchase, Care, and Diseases most frequently encountered, ^\'ith lessons 
for bitting the Horse, and bending the neck. By Bvt. Major A. K. 
Arnold, Capt. Fifth Cavalry, Assistant Instructor of Cavalry Tactics, 
U. S. Mil. Academj. 

Cooke's Cavalry Practice. 

100 Illustrations. 12mo. Cloth. $1.00. 
Cavalry Tactics; Regulations for the Instruction, Formation and 
Movements of the Cavalry of the Army and Volunteers of the United 
States. By Philip St. George Cooke, Brig.-Gen. U. S. A. 
This is the edition now iu use in the U. S. Army. 

Patten's Cavalry Drill. 

93 Engraviugg. 12mo. Paper. 50c. 
Cavalry Drill. Containing Instructions on Foot ; Instructions on 
Horseback ; Basis of Instruction ; School of the Squadron, and Sabre 
Exercise. 



Patten's Infantry Tactics. 

92 Engravings. 12mo. Paper. 50c. 
Infantry Tactics. School of the Soldier ; Manual of Arms for the 
Rifle Musket ; Instructions for Recruits, School of the Company ; 
Skirmishers, or Light Infantry and Rifle Company Movements ; the 
Bayonet Exercise ;' the Small-Sword Exercise ; Manual of the Sword 
or Sabre. 

Patten's Infantry Tactics. 

Revised Edition. 100 Engravings, 12mo. Paper. 75c. 
Infantry Tactics. Contains Nomenclature of the Musket; School 
of the Company ; Skirmishers, or Light Infantry and Rifle Company 
Movements ; School of the Battalion ; Bayonet Exercise ; Small Sword 
Exercise ; Manual of the Sword or Sabre. 



Patten's Army Man^^al. 

8vo. Cloth. S2.00. 
Army Manual. Containing Instructions for Officers in the Preparation 
of Rolls, Returns, and Accounts required of Regimental and Company 
Commanders, and pertaining to the Subsistence and Quartermaster's 
Department, &c., &c. 



Patten's Artillery l>rill.' 

12mo. Papei. 50c. 
Artillery Drill. Containing instruction in the School of the Piece, 
and Battery Manoeuvres, compiled agi-eeably to the Latest Regulations 
of the War Department. From Standard Military Authority. By 
George Patten, late tJ. S. Army. 

AndreAvs' Hints to Company Officers. 

ISrao. Cloth. 60c. 
Hints to Company Officers on their Military Duties. By 
General C. C. Andrews, Third Regt, Minnesota Vols. 



Thomas' Rifled. Ordnance. 

Fifth Editiou, Revised. Illustrated. 8to. Cloth. $2.00. 
Rifled Ordnance ; A Practical Treatise on the Application of the 
Principle of the Rifle to Guns and Mortars of every calibre. To which 
is added a new theory of the initial action and force of Fired Gun- 
powder. By Lynall Thomas, F. R. S. L. 



Brinkerh-ofF's Volunteer Q^iartermaster. 

12mo. Cloth. $2.50. 
The Volunteer Quartermaster. By Captain R. Brinkerholf , Post 
Quartermaster at Washington. 



Hnnter's Manual for Qnartermasters and 

Commissaries. 

12mo. Cloth. $1.25. Flexible Morocco, Si. 50. 
Manual for Quartermasters and Commissaries. Containing 
Instructions in the Preparation of Vouchers, Abstracts, Returns, etc. 
By Captain R. F. Hunter, late of the U. S. Army. 12mo. Cloth. 
$1.25. 



Grreener's Griinnery. 

8vo. Cloth. S4.00. Full Calf. SG.OO. 
Gunnery in 1858. A Treatise on Rifles, Cannon, and Sporting Arms. 
By W'm. Greener, R. C. E. 



Head's System of Fortifications. 

Illustrated. 4to. Paper. $1.00. 
A New System of Fortifications. By George E. Head, A. M., 
Capt. Twenty-Ninth Infantry, and Bvt. Major U. S. A. 






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4* 






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